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.
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