Solveeit Logo

Question

Question: A car moving with speed v on a straight road can be stopped within distance d on applying brakes. If...

A car moving with speed v on a straight road can be stopped within distance d on applying brakes. If the same car is moving with speed 3v and brakes provide half retardation, then the car will stop after travelling distance.

Explanation

Solution

To find the distance at which the car stops, first we need to find the acceleration of the car after the brakes are applied. After finding the acceleration we will find the distance required to stop the car if the acceleration or retardation is halved. The formula used is:
V2=U2+2aS{{V}^{2}}={{U}^{2}}+2aS
where V is the final velocity, U is the initial velocity, a is the acceleration and S is the distance stopped.

Complete step by step answer:
The speed with which the car is moving is given as velocity v.
Now, the car has stopped due to the application of the brakes making the final velocity zero.
Hence, the distance to stoppage is d and the initial velocity is v making the acceleration of the car till stoppage as a, and to find the value of "a", we use the formula of V2=U2+2aS{{V}^{2}}={{U}^{2}}+2aS.
After finding the acceleration, we place the value of the acceleration in the second case or the case when the car comes to stop and as given in the question, the car's acceleration is reduced to half hence, the acceleration when the car retards after braking is a2\dfrac{a}{2}. Now the velocity with which the car moves after retardation is 3v3v and acceleration is a=v24da=\dfrac{{{v}^{2}}}{4d}. Placing all the values in the distance formula we get the value of the distance covered after braking is:
V2=U2+2aS{{V}^{2}}={{U}^{2}}+2aS
Placing the value of v as 3v3v and a as v24d\dfrac{{{v}^{2}}}{4d}, we get the distance as:
S=(3v)22(v24d)S=\dfrac{{{(3v)}^{2}}}{2\left( \dfrac{{{v}^{2}}}{4d} \right)}
S = 18dS\text{ }=\text{ }18d
Therefore, after braking the car will stop at a distance of 18d18d.

Note:
The term retardation means backward or negative acceleration where the acceleration is used during braking purposes, so the retardation is halved meaning the acceleration becomes half of that of the acceleration when the car is moving forward.