Question

that's physics question. A truck and a car are moving on smooth, level road such that the kinetic energy associated with them is same. breaks are applied simultaneously in both of them such that equal retarding forces are produced in them. which one will cover a greater distance,before it stop?

Answer 1

Obviously the car will travel more distance.

As K.E are same


velocity of car will be greater.

Answer 2

The Car will cover more Distance.

Reason:

K.E. of Truck = K.E. of Car (Given)

$\frac{1}{2} m_{1} v_{1} ^{2} = \frac{1}{2} m_{2} v_{2} ^{2}$

$m_{1} v_{1} ^{2} = m_{2} v_{2} ^{2}$

Here, Mass of Truck > Mass of Car

$m_{1} > m_{2}$

As, Mass of Truck is greater, So Velocity of Truck must be less and Mass of Car is less, So Velocity of Car should be more to equalise each other.

Now, We know

$v = \frac{s}{t}$

Therefore,

$v \: \: \alpha \: \: s$

Hence, As velocity of Car is more, then Distance travelled by it is more.