Finding acceleration of a car when it is braking

AI Thread Summary
To find the acceleration of a car braking from 20 m/s to a stop over 50 meters, the correct approach involves using the SUVAT equations rather than the time formula for constant speed. The acceleration can be calculated using the equation: a = (final velocity² - initial velocity²) / (2 * distance), which yields an acceleration of -4 m/s², indicating deceleration. The force applied by the braking system can then be determined using Newton's second law, F = ma, where m is the mass of the car. Understanding that braking involves deceleration is crucial for solving these types of problems. Proper application of the SUVAT equations is essential for accurate calculations in kinematics.
Amelina Yoo
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Homework Statement


A 1200kg car moving at 20 metres per second stops in a distance of 50m when the brakes are applied.
a) What is the acceleration of the car when it is braking?
b) Find the force applied by the car's braking system.

Homework Equations


For a) I used the acceleration formula: a=(velocity - initial velocity)/t
Time formula from rearranging: t=s/v

The Attempt at a Solution


a)
t=s/v
t=50/20=2.5s

velocity=50/2.5=20 metres per second

a=(20-0)/2.5 = 8ms(^-2)[/B]
I do not know how to find the acceleration when BRAKING, and the answer claims that it is 4ms(^-2.)

As for b, I believe that I am unable to attempt it seeing as I am unable to find the acceleration.

Please explain how to do it! I have read through the chapter of my textbook, and am unable to understand. thank you very much!
 
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Amelina Yoo said:

Homework Statement


A 1200kg car moving at 20 metres per second stops in a distance of 50m when the brakes are applied.
a) What is the acceleration of the car when it is braking?
b) Find the force applied by the car's braking system.

Homework Equations


For a) I used the acceleration formula: a=(velocity - initial velocity)/t
Time formula from rearranging: t=s/v

The Attempt at a Solution


a)
t=s/v
t=50/20=2n5s

velocity=50/2.5=20 metres per second

a=(20-0)/2.5 = 8ms(^-2)[/B]
I do not know how to find the acceleration when BRAKING, and the answer claims that it is 4ms(^-2.)

As for b, I believe that I am unable to attempt it seeing as I am unable to find the acceleration.

Please explain how to do it! I have read through the chapter of my textbook, and am unable to understand. thank you very much!

Braking to a stop is just the reverse of accelerating to speed from a standing start.

Your equation t = s/v works only for constant speed. I have no idea what "2n5s" means.

There is a SUVAT equation which relates acceleration, distance, and initial and final velocities which would be rather useful here.
 
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