Question

The brakes of an unloaded lorry of mass 1000 kg will slow it down from 40 to 20 km/h in 7.5 s. How long will they take to stop it from a speed of 30 km/h, if it has taken on a load of
2200 kg?

Answer 1

Step 1: Given data

Mass of lorry is, $m_1=1000kg$

Initial velocity is, $u=40km/h$

Final velocity is, $v=20km/hr$

Time taken to slow down from the initial to final velocity is, $t=7.5s$

Mass of load taken by the truck is, $m_2=2200kg$

Step 2: Calculating force

Force is given as,

$F=ma$

where, $m$ is mass and $a$ is acceleration.

Acceleration is given as,

$a=\frac{\Delta v}{\Delta t}$

where, $\Delta v$ is change in velocity and $\Delta t$ is change in time

Therefore, the equation for force can be written as,

$F=\frac{m\Delta v}{\Delta t}$

The mass of the lorry will be considered here, therefore, the force will be calculated as:

$F=\frac{m_1(u-v)}{t}$

Substituting the known values in the above equation, we get,

$$\begin{gathered} F=\frac{1000kg\times (40-20)km/hr}{7.5s} \\ F=\frac{1000\times 20}{7.5} \\ F=26666.67N \end{gathered}$$

Step 3: Calculating time taken to stop when the speed of the truck is $30km/hr$

Now,

Initial velocity is, $u=30km/hr$

Final velocity is, $v=0km/hr$

New mass will be,

$$\begin{gathered} m=m_1+m_2 \\ m=1000kg+2200kg \\ m=3200kg \end{gathered}$$

The relationship between force and velocity can be written as,

$F=\frac{m(u-v)}{t}$

Substituting the known values, we get,

$$\begin{gathered} 26666.67N=\frac{3200kg\times (30-0)km/hr}{t} \\ t=\frac{3200\times 30}{2666.67} \\ t=36s \end{gathered}$$

Therefore, the time taken will be $36s$.