A motorist traveling at 10m/s, was able to bring his car to rest in a distance of 10m. If he had been traveling at 30m/s, in what distance could he bring his cart to rest using the same breaking force?

By just a quick look you might be tempted to say 30 m. However the key information is that the breaking force is the same. We can calculate the deceleration of the car when at 10m/s using the equation of motion:

v2=u2+2as (1),

where u is the initial velocity= 10 m/s, v the final velocity which is zero since it stops, s displacement and a acceleration. By substituting the values you end up with an acceleration a= -5 m/s2. In order to find the force of the car, we use the equation

F=ma (Newton's Second Law) (2),

where F is the  breaking force, m is the mass of the car and a is the acceleration(here deceleration). Thus, F= 5m N. However since the mass of the car doesn't change when it travels at 30 m/s and the force is the same, deceleration is the same too. Using the same equation of motion (1), with values u=30m/s, v=0m/s, a=5m/s2 we find that s=90 m which is the distance travelled before the car comes to a stop.

CT
Answered by Charis T. Physics tutor

21083 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

From what height, h, should a rail-cart fall to complete a loop-the-loop of radius r without falling off a the track? Assume the track on which the rail-cart travels is smooth and express h in terms of r.


How can I describe the motion of an object falling, due to gravity, through a fluid? And when does the object reach terminal velocity?


Two trains are heading in opposite directions on the same track. Train X has a mass of 16000kg and a speed of 2.8m/s. Train Y has a mass of 12000kg and a speed of 3.1m/s. At what speed do the joined trains move off together immediately after the collison?


What is the escape velocity of an object leaving a planet mass M, radius R?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning