Question on Equation Of Motion

Hello,

Could anyone please help me with this question.

Two cars are travelling towards each other on a single-track road with equal speeds of 35m/s. When they are a distance of 500m apart, they both decide to brake.

(a) What minimum equal decelerations would they require just to avoid an acceident? The brake on one car fail and it continues with the same speed. The other car slows down and at the point of collision it has just stopped.

(b)What distances have the two cars travelled when the collision occurs?
(c)What time has elapsed?

My attempt:

a) So the point of collision is when both car covers 250m. It will take 250/35= 7.142857.... seconds.

initial velocity=35m/s
final velocity=0m/s
time=7.142857
distance=250m

Applying motion equation: V^2=u^2+2as

0=1225+2x250s, 500s=-1225
s=-2.45 m/s^2 Thus deceleration is 2.45m/s^2

(b) I spent alot of time on this but can't visualize the question.
initial velocity=35m/s
final velocity=?
time=?
distance=?

second car

initial velocity=35m/s
final velocity=?
a=-2.45m/s^2
distance=?

using V^2=u^2+2as (car 1)
V^2=1225+2as

Please help. this question is confusing me. Question says "the other car slows down and at the point of collision it has just stopped.(Does this mean they don't crash? I think decelerating car will not stop after 250m because the other car is coming toward without decreasing the speeed so collision will occur somewhere near the decelerating car side.

Thanks for your help.

Reply 1

username9816

14

charikaar

Could anyone please help me with this question.

Two cars are travelling towards each other on a single-track road with equal speeds of 35m/s. When they are a distance of 500m apart, they both decide to brake.

(a) What minimum equal decelerations would they require just to avoid an acceident?

The brake on one car fail and it continues with the same speed. The other car slows down and at the point of collision it has just stopped.

(b)What distances have the two cars travelled when the collision occurs?
(c)What time has elapsed?

a.) To just avoid an accident and for both cars to have equal decelerations and hence equal distances travelled before stopping (as initial speed is the same), both cars would need to travel 250m.

:. for either car: u = 35, a = a, s = 250, v = 0
v^2 = u^2 + 2as
-> 0 = 1225 + 500a
-> a = -(1225)/500
-> a = - 2.45 ms^-2
-> Min. Deceleration Required = 2.45 ms^-2

b.) 250m; if the car decelerating just reaches v = 0 as the collision occurs, this happens when it has travelled 250m.

c.) Well for the decelerating car: u = 35, a = -2.45, s = 250, v = 0, t = t.
v = u + at
-> 0 = 35 - 2.45t
-> 2.45t = 35
-> t = 14.3 s
-> Time Elapsed (POC) = 14.3 s

Reply 2

Eddie K

2

Do they give a decelleration to use for Car 1? I do not want to assume they are want us to use the answer from part a), which you have correctly calculated, even though you have gone round the houses!

Reply 3

Eddie K

2

Nima's calculation is correct, using a=-2.45m/s^2 for part b).

I did not want to assume that was the still the correct decelleration!

Reply 4

username9816

14

Eddie K

I did not want to assume that was the still the correct decelleration!

You use that answer in part b, surely? If you didn't, how would you calculate the distance travelled when collision occurs; this is dependant on the deceleration of the car that is braking. We calculated that above.

Reply 5

charikaar

OP

Nima

a.) To just avoid an accident and for both cars to have equal decelerations and hence equal distances travelled before stopping (as initial speed is the same), both cars would need to travel 250m.

:. for either car: u = 35, a = a, s = 250, v = 0
v^2 = u^2 + 2as
-> 0 = 1225 + 500a
-> a = -(1225)/500
-> a = - 2.45 ms^-2
-> Min. Deceleration Required = 2.45 ms^-2

b.) 250m; if the car decelerating just reaches v = 0 as the collision occurs, this happens when it has travelled 250m.

c.) Well for the decelerating car: u = 35, a = -2.45, s = 250, v = 0, t = t.
v = u + at
-> 0 = 35 - 2.45t
-> 2.45t = 35
-> t = 14.3 s
-> Time Elapsed (POC) = 14.3 s

Thanks for your help. I don't have the answer sheet so I won't know if your answer to part b is correct. Using common sense, wouldn't non decelerating car travel more than 250 meters so collision occrus somewhere nearer decelerating car side.

Charikaar

Reply 6

charikaar

OP

Thanks. I got it now. The question does say that when the collision occurs, this car stopped (i.e. had a velocity of 0). I better ignore everything else.

Charikaar

Reply 7

Eddie K

2

Nima

You use that answer in part b, surely? If you didn't, how would you calculate the distance travelled when collision occurs; this is dependant on the deceleration of the car that is braking. We calculated that above.

Yes, but no where in the question does it suggest that it is the same situation. I don't like assuming things, so I thought I'd ask before I started doing the question!

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