Question

A truck of mass 9000kg is moving with speed 18m/s, when the driver decides to stop and applies the brakes. After 6s the truck stops. Assuming that the stopping is with constant deceleration calculate the followin:

a)The distance traveled by the car during stopping
b)The acceleration of the truck
c)The loss of energy due to friction

Answer 1

Answer on Question#56401 - Physics - Mechanics - Relativity

A truck of mass $m = 9000 \, \text{kg}$ is moving with speed $v_i = 18 \, \frac{\text{m}}{\text{s}}$, when the driver decides to stop and applies the brakes. After $t = 6 \, \text{s}$ the truck stops. Assuming that the stopping is with constant deceleration calculate the following:

a) The distance traveled by the car during stopping

b) The acceleration of the truck

c) The loss of energy due to friction

Solution:

Let the acceleration of the truck be $a$, then the final speed $v_f$ of the truck and the deceleration time $t$ are related by

$$v_f - v_i = a t$$

Since $v_f = 0 \, \frac{\text{m}}{\text{s}}$, we obtain

$$a = - \frac {v_i}{t} = - \frac {18 \, \frac{\text{m}}{\text{s}}}{6 \, \text{s}} = - 3 \, \frac{\text{m}}{\text{s}^2}$$

The distance $l$ traveled by truck is

$$l = v_i t + \frac {a t^2}{2} = 18 \, \frac{\text{m}}{\text{s}} \cdot 6 \, \text{s} - \frac {3 \, \frac{\text{m}}{\text{s}^2} (6 \, \text{s})^2}{2} = 54 \, \text{m}$$

According to the law of conservation of energy the loss of energy is equal to the initial kinetic energy of the truck

$$E_l = \frac {m v_l^2}{2} = \frac {9000 \, \text{kg} \left(18 \, \frac{\text{m}}{\text{s}}\right)^2}{2} = 1458 \, \text{kJ}$$

Answer:

a) $54 \, \text{m}$

b) $-3 \, \frac{\text{m}}{\text{s}^2}$

c) $1458 \, \text{kJ}$