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Hi, I need help with the following questions:

A lion is watching a zebra from 35 m behind it. Both are stationary. The lion then starts chasing by accelerating at a constant rate of 3 m s^-2 for 5 s. Once at top speed the lion decelerates at 0.5 m s^-2. The zebra starts moving Is after the lion started, accelerating at a constant rate of 2 m s^-2 for 7s before maintaining a constant speed.

a) Show that the lion has not caught the zebra after 8 s.

b) Show that the gap between them at time ts, for t > 8, after the start of the lion’s motion is given by 1/4 t^2 - 5t + 51/2 and, hence, determine when the lion catches the zebra, or when the lion gets closest and how close it gets.

A car is behind a tractor on a single-lane straight road with one lane in each direction. Both are moving at 15ms-1. The speed limit is 25 m s^-1, so the car wants to overtake. The safe distance between the car and the tractor is 20 m.

a) To overtake, the car goes onto the other side of the road and accelerates at a constant rate of 2 m s^-2 until reaching the speed limit, when it continues at constant speed. Show that the distance the car is ahead of the tractor at time ts after it starts to accelerate is given by t^2 - 20 for 0 < t < 5, and deduce that the car is not a safe distance ahead of the tractor before reaching the speed limit.

b) The car pulls in ahead of the tractor once it is a safe distance ahead. Find the total time taken from the start of the overtaking manoeuvre until the car has safely overtaken the tractor.

c) To overtake safely on the single-lane road, when the car returns to the correct side of the road in front of the tractor there must be a gap between the car and oncoming traffic of at least 20 m. Assuming a car travelling in the opposite direction is moving at the speed limit, find the minimum distance it must be from the initial position of the overtaking car at the point at which it starts to overtake.

For Question 1, I need help with part b only. WHile for Question 2, I need help with part c only. Thank you.

**Question 1**A lion is watching a zebra from 35 m behind it. Both are stationary. The lion then starts chasing by accelerating at a constant rate of 3 m s^-2 for 5 s. Once at top speed the lion decelerates at 0.5 m s^-2. The zebra starts moving Is after the lion started, accelerating at a constant rate of 2 m s^-2 for 7s before maintaining a constant speed.

a) Show that the lion has not caught the zebra after 8 s.

b) Show that the gap between them at time ts, for t > 8, after the start of the lion’s motion is given by 1/4 t^2 - 5t + 51/2 and, hence, determine when the lion catches the zebra, or when the lion gets closest and how close it gets.

**Question 2**A car is behind a tractor on a single-lane straight road with one lane in each direction. Both are moving at 15ms-1. The speed limit is 25 m s^-1, so the car wants to overtake. The safe distance between the car and the tractor is 20 m.

a) To overtake, the car goes onto the other side of the road and accelerates at a constant rate of 2 m s^-2 until reaching the speed limit, when it continues at constant speed. Show that the distance the car is ahead of the tractor at time ts after it starts to accelerate is given by t^2 - 20 for 0 < t < 5, and deduce that the car is not a safe distance ahead of the tractor before reaching the speed limit.

b) The car pulls in ahead of the tractor once it is a safe distance ahead. Find the total time taken from the start of the overtaking manoeuvre until the car has safely overtaken the tractor.

c) To overtake safely on the single-lane road, when the car returns to the correct side of the road in front of the tractor there must be a gap between the car and oncoming traffic of at least 20 m. Assuming a car travelling in the opposite direction is moving at the speed limit, find the minimum distance it must be from the initial position of the overtaking car at the point at which it starts to overtake.

For Question 1, I need help with part b only. WHile for Question 2, I need help with part c only. Thank you.

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#2

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Hi, I need help with the following questions:

A lion is watching a zebra from 35 m behind it. Both are stationary. The lion then starts chasing by accelerating at a constant rate of 3 m s^-2 for 5 s. Once at top speed the lion decelerates at 0.5 m s^-2. The zebra starts moving Is after the lion started, accelerating at a constant rate of 2 m s^-2 for 7s before maintaining a constant speed.

a) Show that the lion has not caught the zebra after 8 s.

b) Show that the gap between them at time ts, for t > 8, after the start of the lion’s motion is given by 1/4 t^2 - 5t + 51/2 and, hence, determine when the lion catches the zebra, or when the lion gets closest and how close it gets.

A car is behind a tractor on a single-lane straight road with one lane in each direction. Both are moving at 15ms-1. The speed limit is 25 m s^-1, so the car wants to overtake. The safe distance between the car and the tractor is 20 m.

a) To overtake, the car goes onto the other side of the road and accelerates at a constant rate of 2 m s^-2 until reaching the speed limit, when it continues at constant speed. Show that the distance the car is ahead of the tractor at time ts after it starts to accelerate is given by t^2 - 20 for 0 < t < 5, and deduce that the car is not a safe distance ahead of the tractor before reaching the speed limit.

b) The car pulls in ahead of the tractor once it is a safe distance ahead. Find the total time taken from the start of the overtaking manoeuvre until the car has safely overtaken the tractor.

c) To overtake safely on the single-lane road, when the car returns to the correct side of the road in front of the tractor there must be a gap between the car and oncoming traffic of at least 20 m. Assuming a car travelling in the opposite direction is moving at the speed limit, find the minimum distance it must be from the initial position of the overtaking car at the point at which it starts to overtake.

For Question 1, I need help with part b only. WHile for Question 2, I need help with part c only. Thank you.

**ewchiew**)Hi, I need help with the following questions:

**Question 1**A lion is watching a zebra from 35 m behind it. Both are stationary. The lion then starts chasing by accelerating at a constant rate of 3 m s^-2 for 5 s. Once at top speed the lion decelerates at 0.5 m s^-2. The zebra starts moving Is after the lion started, accelerating at a constant rate of 2 m s^-2 for 7s before maintaining a constant speed.

a) Show that the lion has not caught the zebra after 8 s.

b) Show that the gap between them at time ts, for t > 8, after the start of the lion’s motion is given by 1/4 t^2 - 5t + 51/2 and, hence, determine when the lion catches the zebra, or when the lion gets closest and how close it gets.

**Question 2**A car is behind a tractor on a single-lane straight road with one lane in each direction. Both are moving at 15ms-1. The speed limit is 25 m s^-1, so the car wants to overtake. The safe distance between the car and the tractor is 20 m.

a) To overtake, the car goes onto the other side of the road and accelerates at a constant rate of 2 m s^-2 until reaching the speed limit, when it continues at constant speed. Show that the distance the car is ahead of the tractor at time ts after it starts to accelerate is given by t^2 - 20 for 0 < t < 5, and deduce that the car is not a safe distance ahead of the tractor before reaching the speed limit.

b) The car pulls in ahead of the tractor once it is a safe distance ahead. Find the total time taken from the start of the overtaking manoeuvre until the car has safely overtaken the tractor.

c) To overtake safely on the single-lane road, when the car returns to the correct side of the road in front of the tractor there must be a gap between the car and oncoming traffic of at least 20 m. Assuming a car travelling in the opposite direction is moving at the speed limit, find the minimum distance it must be from the initial position of the overtaking car at the point at which it starts to overtake.

For Question 1, I need help with part b only. WHile for Question 2, I need help with part c only. Thank you.

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(Original post by

part of the rules of this forum is that we cannot give full solutions... maybe just post your working out and tell us where you're stuck

**Kundana Amudala**)part of the rules of this forum is that we cannot give full solutions... maybe just post your working out and tell us where you're stuck

When t>8, the displacement of lion is s

_{L}=75/2 + 3(t-5) - 1/4(t-5)

^{2}, while the displacement of zebra is s

_{Z}=84 + 14(t-8). This is what I got.

While for Question 2(c), I have no idea how to start. Can you please guide me through the steps of solving this question?

Thank you.

Last edited by ewchiew; 2 months ago

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#4

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For Question 1(b), I just want to know how to find the displacement of the lion and the zebra, then the remaining parts I can do it. I tried to find the answer, but it looks weird. I do not know whether my understanding of the situation is correct or not.

When t>8, the displacement of lion is s

While for Question 2(c), I have no idea how to start. Can you please guide me through the steps of solving this question?

Thank you.

**ewchiew**)For Question 1(b), I just want to know how to find the displacement of the lion and the zebra, then the remaining parts I can do it. I tried to find the answer, but it looks weird. I do not know whether my understanding of the situation is correct or not.

When t>8, the displacement of lion is s

_{L}=75/2 + 3(t-5) - 1/4(t-5)^{2}, while the displacement of zebra is s_{Z}=84 + 14(t-8). This is what I got.While for Question 2(c), I have no idea how to start. Can you please guide me through the steps of solving this question?

Thank you.

to show the lion never reaches the zebra, show the displacement of the zebra relative to the lion is always positive,

Last edited by mqb2766; 2 months ago

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(Original post by

can you upload your working?

to show the lion never reaches the zebra, show the displacement of the zebra relative to the lion is always positive,

**mqb2766**)can you upload your working?

to show the lion never reaches the zebra, show the displacement of the zebra relative to the lion is always positive,

For part (a), I actually calculate separately and minus the two displacements. I got 80.25 m for Lion and 84 m for zebra.

Now what I want to know is part (b). And my formulation are shown above, but I not sure whether it is correct.

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#6

(Original post by

Yeah, my velocity functions for both animals are in piecewise form (two functions in one).

For part (a), I actually calculate separately and minus the two displacements. I got 80.25 m for Lion and 84 m for zebra.

Now what I want to know is part (b). And my formulation are shown above, but I not sure whether it is correct.

**ewchiew**)Yeah, my velocity functions for both animals are in piecewise form (two functions in one).

For part (a), I actually calculate separately and minus the two displacements. I got 80.25 m for Lion and 84 m for zebra.

Now what I want to know is part (b). And my formulation are shown above, but I not sure whether it is correct.

Last edited by mqb2766; 2 months ago

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(Original post by

Can you upload your actual workings for both a) and b)?

**mqb2766**)Can you upload your actual workings for both a) and b)?

v(Zebra) = 2(t - 1). 1<=t<=8 ; 14, t>8

At t=8, s(Lion) = 1/2(0+15)(5) + 1/2(15+13.5)(3) = 80.25 m

s(Zebra) = 35 + 1/2(0+14)(7) = 84 m

Since s(Zebra) > s(Lion), so the lion has not caught the zebra.

b) For t>8 , s(Lion) = 1/2(3)(5)^2 + 3(t - 5) + 1/2(-1/2)(t - 5)^2 = 75/2 + 3(t - 5) - 1/4(t - 5)2

s(Zebra) = 84 + 14(t - 8) = 14t - 28

I have just done until here. Correct me if I am wrong. Thanks.

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#8

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a) v(Lion) = 3t, 0<=t<=5 ; 15 - 1/2(t - 5), t>5

v(Zebra) = 2(t - 1). 1<=t<=8 ; 14, t>8

At t=8, s(Lion) = 1/2(0+15)(5) + 1/2(15+13.5)(3) = 80.25 m

s(Zebra) = 35 + 1/2(0+14)(7) = 84 m

Since s(Zebra) > s(Lion), so the lion has not caught the zebra.

b) For t>8 , s(Lion) = 1/2(3)(5)^2 + 3(t - 5) + 1/2(-1/2)(t - 5)^2 = 75/2 + 3(t - 5) - 1/4(t - 5)2

s(Zebra) = 84 + 14(t - 8) = 14t - 28

I have just done until here. Correct me if I am wrong. Thanks.

**ewchiew**)a) v(Lion) = 3t, 0<=t<=5 ; 15 - 1/2(t - 5), t>5

v(Zebra) = 2(t - 1). 1<=t<=8 ; 14, t>8

At t=8, s(Lion) = 1/2(0+15)(5) + 1/2(15+13.5)(3) = 80.25 m

s(Zebra) = 35 + 1/2(0+14)(7) = 84 m

Since s(Zebra) > s(Lion), so the lion has not caught the zebra.

b) For t>8 , s(Lion) = 1/2(3)(5)^2 + 3(t - 5) + 1/2(-1/2)(t - 5)^2 = 75/2 + 3(t - 5) - 1/4(t - 5)2

s(Zebra) = 84 + 14(t - 8) = 14t - 28

I have just done until here. Correct me if I am wrong. Thanks.

Its worth noting, I don't get the given answer. I suspect the question has been edited incorrectly.

Last edited by mqb2766; 2 months ago

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(Original post by

A bit of explanation would help explain what you're trying to do, but why the 3(t-5) for the lion t>8?

Its worth noting, I don't get the given answer. I suspect the question has been edited incorrectly.

**mqb2766**)A bit of explanation would help explain what you're trying to do, but why the 3(t-5) for the lion t>8?

Its worth noting, I don't get the given answer. I suspect the question has been edited incorrectly.

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#10

(Original post by

Sorry, my fault. It should be 15(t-5) .

**ewchiew**)Sorry, my fault. It should be 15(t-5) .

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(Original post by

I agree. If you expand your quadratic zebra - lion displacement now, you get something like the desired solution, but not equal to it. The desired solution has the lion getting closer to the zebra at t=8, but this isn't true as zebra = 14m/s and lion = 13.5m/s. However, just use your solution and show the lion never catches the zebra and find the closest approach.

**mqb2766**)I agree. If you expand your quadratic zebra - lion displacement now, you get something like the desired solution, but not equal to it. The desired solution has the lion getting closer to the zebra at t=8, but this isn't true as zebra = 14m/s and lion = 13.5m/s. However, just use your solution and show the lion never catches the zebra and find the closest approach.

But the problem is when I minus the two displacements, I do not get the one as shown (1/4 t^2 - 5t + 51/2).

And if I used mine one to solve, I able to solve for values of t, and the value of t is less than 8.

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#12

(Original post by

Actually, I know this step.

But the problem is when I minus the two displacements, I do not get the one as shown (1/4 t^2 - 5t + 51/2).

And if I used mine one to solve, I able to solve for values of t, and the value of t is less than 8.

**ewchiew**)Actually, I know this step.

But the problem is when I minus the two displacements, I do not get the one as shown (1/4 t^2 - 5t + 51/2).

And if I used mine one to solve, I able to solve for values of t, and the value of t is less than 8.

Last edited by mqb2766; 2 months ago

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#13

If you want help with the other question, upload what you;ve done for the other parts?

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(Original post by

Third time, the given answer does not match the question information and I agree about the closest approach being for t<8 so the quadratic not being valid. An alternative way to say this, is to say the closest approach for the quadratic is at t=8.

**mqb2766**)Third time, the given answer does not match the question information and I agree about the closest approach being for t<8 so the quadratic not being valid. An alternative way to say this, is to say the closest approach for the quadratic is at t=8.

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(Original post by

If you want help with the other question, upload what you;ve done for the other parts?

**mqb2766**)If you want help with the other question, upload what you;ve done for the other parts?

But for part c, I am unclear with the question, so I do not know how to start.

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#16

(Original post by

Part a and b I had done it correctly.

But for part c, I am unclear with the question, so I do not know how to start.

**ewchiew**)Part a and b I had done it correctly.

But for part c, I am unclear with the question, so I do not know how to start.

A sketch would help with c) but it is again about relative velocity / displacement, the only real difference is that the car in the other lane would be moving with negative velocity / towards the original car? You'd be using parts a and b) and applying the overtaking time (part b) to the motion of the car in the other lane.

Last edited by mqb2766; 2 months ago

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(Original post by

Can you upload what you did for the other two parts pls.

A sketch would help with c) but it is again about relative velocity / displacement, the only real difference is that the car in the other lane would be moving with negative velocity / towards the original car? You'd be using parts a and b) and applying the overtaking time (part b) to the motion of the car in the other lane.

**mqb2766**)Can you upload what you did for the other two parts pls.

A sketch would help with c) but it is again about relative velocity / displacement, the only real difference is that the car in the other lane would be moving with negative velocity / towards the original car? You'd be using parts a and b) and applying the overtaking time (part b) to the motion of the car in the other lane.

D = s(car) - s(tractor) = (t^2 + 15t -20) - (15t) = t^2 -20

When t=5 , D = (5)^2 - 20 = 5 m (Car is ahead the tractor by 5 m (< 20 m), so there is no safe distance).

b) At t = 5, D = 5 m

The car needs to ahead by another 15 m to reach the safe distance.

After t = 5, s(car) = 25t , s(tractor) = 15t

New D = 25t - 15t = 10t

D = 15 , 10t = 15 ---> t = 1.5

The total time taken = 5 s + 1.5 s = 6.5 s

This is my solution for the first two parts.

For part c, it mentioned "from the initial position of the overtaking car at the point at which it starts to overtake". Do I need to find the time and displacement when the same lane car started to overtake the tractor? But then how can I formulate the displacement of the opposite lane car?

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#18

(Original post by

a) s(car) = t^2 + 15t -20 , s(tractor) = 15t for 0<=t<=5

D = s(car) - s(tractor) = (t^2 + 15t -20) - (15t) = t^2 -20

When t=5 , D = (5)^2 - 20 = 5 m (Car is ahead the tractor by 5 m (< 20 m), so there is no safe distance).

b) At t = 5, D = 5 m

The car needs to ahead by another 15 m to reach the safe distance.

After t = 5, s(car) = 25t , s(tractor) = 15t

New D = 25t - 15t = 10t

D = 15 , 10t = 15 ---> t = 1.5

The total time taken = 5 s + 1.5 s = 6.5 s

This is my solution for the first two parts.

For part c, it mentioned "from the initial position of the overtaking car at the point at which it starts to overtake". Do I need to find the time and displacement when the same lane car started to overtake the tractor? But then how can I formulate the displacement of the opposite lane car?

**ewchiew**)a) s(car) = t^2 + 15t -20 , s(tractor) = 15t for 0<=t<=5

D = s(car) - s(tractor) = (t^2 + 15t -20) - (15t) = t^2 -20

When t=5 , D = (5)^2 - 20 = 5 m (Car is ahead the tractor by 5 m (< 20 m), so there is no safe distance).

b) At t = 5, D = 5 m

The car needs to ahead by another 15 m to reach the safe distance.

After t = 5, s(car) = 25t , s(tractor) = 15t

New D = 25t - 15t = 10t

D = 15 , 10t = 15 ---> t = 1.5

The total time taken = 5 s + 1.5 s = 6.5 s

This is my solution for the first two parts.

For part c, it mentioned "from the initial position of the overtaking car at the point at which it starts to overtake". Do I need to find the time and displacement when the same lane car started to overtake the tractor? But then how can I formulate the displacement of the opposite lane car?

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(Original post by

I presume its when the original car pulls out into the lane. You've worked out how long the manoeuvre takes, so finding the distance travelled by the 2nd car should be straightforward? A sketch of what you're thinking may help if you're unsure.

**mqb2766**)I presume its when the original car pulls out into the lane. You've worked out how long the manoeuvre takes, so finding the distance travelled by the 2nd car should be straightforward? A sketch of what you're thinking may help if you're unsure.

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#20

(Original post by

Sorry, I still do not get your idea. Can you please explain further? Thanks.

**ewchiew**)Sorry, I still do not get your idea. Can you please explain further? Thanks.

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