Answer to Question #287368 in Mechanics | Relativity for lau

Question

A lorry of mass 10Mg reaches a speed of 63km/h in 20s, starting from rest with uniform acceleration. It travels 2km at this speed and is then brought to rest in a distance of 125m.

Determine: (a) the momentum at 63km/h

(b) the accelerating and braking forces required.

Answer 1

Explanations & calculations

• $\small 10\,\text{Mg}= 10000\, kg$
• $\small 63\, kmh^{-1}= 17.5\,ms^{-1}$

a) Momentum is $\small mv=10^4\,kg\times17.5\,ms^{-1}=1.75\times10^5\,kgms^{-1}$

b)

Acceleration is

$\qquad\quad \begin{aligned} \small v&=\small u+at\\ \small a&=\small \frac{v}{t}=\frac{17.5\,ms^{-1}}{20\,s} \end{aligned}$ By simplifying you can get the answer.

Retarding force required is

$\qquad\quad \begin{aligned} \small K.E&=\small fs\\ \small \frac{1}{2}mv^2 &=\small fs\\ f&=\small \frac{mv^2}{2s} = \frac{(10^4).(17.5^2)}{2(125\,m)}\\ \small \end{aligned}$ Upon simplification you get the answer.