This discussion is now closed.

Check out other Related discussions

- Projectile on a string
- A-Level Physics Mechanics help - "tennis ball"
- A1/A2 Mechanics - Pulleys MadasMaths Question
- Projectile on a string physics question
- m1 mechanics as vectors 2018 internation edexcel q6
- A Level Mechanics.
- Mechanics A Level maths question
- Tension in a Simple Pendulum
- acceleration on an angle
- AS level Mechanics help-pulleys
- A-level Maths Mechanics: SUVAT Question HELP
- Projecttiles
- T Shaped Pendulum
- M2 projectile Q help
- Trajectory of a ball to a moving vehicle
- the issac physics astronomy question
- Physics question
- M1 Projectiles
- Simple harmonic motion question
- A Level Physics Mechanics HELPP!!!

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.

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.

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?

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

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.

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

:. 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

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!

- Projectile on a string
- A-Level Physics Mechanics help - "tennis ball"
- A1/A2 Mechanics - Pulleys MadasMaths Question
- Projectile on a string physics question
- m1 mechanics as vectors 2018 internation edexcel q6
- A Level Mechanics.
- Mechanics A Level maths question
- Tension in a Simple Pendulum
- acceleration on an angle
- AS level Mechanics help-pulleys
- A-level Maths Mechanics: SUVAT Question HELP
- Projecttiles
- T Shaped Pendulum
- M2 projectile Q help
- Trajectory of a ball to a moving vehicle
- the issac physics astronomy question
- Physics question
- M1 Projectiles
- Simple harmonic motion question
- A Level Physics Mechanics HELPP!!!

Latest

Trending

Last reply 1 week ago

Gravitational potential and gravitational potential energy definitions, PhysicsPhysics

3

2

Last reply 2 weeks ago

Physics Paper 1 - Deformation topic - multiple choice question solution?Physics

2

4