# Question #d0f95

Jul 13, 2016

A truck and a car are going with the same velocity $u$. Assuming that it is constant velocity for both and that mass of truck is greater than the mass of the car.
Clearly momentum of the truck is greater than the momentum of the car
Because Momentum$= m a s s \times v e l o c i t y$

Now brakes are applied at the same time.
Implies both the truck and the car are decelerated or retarded at the same time. The deceleration produced in each is given by Newton's second Law
$F = m \times a$
$\implies a = \frac{F}{m}$ ......(1)
We know from kinematics that final velocity $v$ and distance $s$ traveled by an object under constant acceleration $a$ are connected through the expression
${v}^{2} - {u}^{2} = 2 a s$
Inserting value of $a$ from (1) and putting value of $v = 0$, we get
$- {u}^{2} = 2 \left(- \frac{F}{m}\right) s$
$- v e$ sign of acceleration signifies that it is deceleration.

Solving for distance we get
$s = \frac{m}{2 F} {u}^{2}$
And as per initial conditions
$s \propto \frac{m}{F}$

From this relationship we can say that for the same retarding force $F$ car will move less.

However, we need to specify specific values for the respective masses and braking powers to assess the situation.