Question

A car starts from rest and moves with a constant acceleration of $5 \,m / s ^{2}$ for $10 \,s$ before the driver applies the brake. It then decelerates for $5 \,s$ before coming to rest, then the average speed of the car over the entire journey of the car is

• A. 23 m / s
• B. 30 m / s
• C. 33 m / s
• D. 25 m / s

Answer 1

Answer: (D)

Schematic diagram for motion of body is

Average speed $=\frac{\text { Total displacement }}{\text { Total time }}$
In interval $t=0$ to $t=10 s$;
Displacement, $s_{1}=\frac{1}{2} a_{1} t^{2}=\frac{1}{2} \times 5 \times(10)^{2}=250 \,m$
Final velocity, $v_{1}=u+a_{1} t$
$\Rightarrow \, v_{1}=0+5 \times 10=50 \,ms ^{-1}$
In interval $t=10 \,s$ to $t=15\, s$,
$v_{1}=50 \,ms ^{-1}, v_{2}=0 \,ms ^{-1}, t=5s$
Using $v=u+at$
We have, $ a_{2}=\frac{0-50}{5}=-10 \,ms ^{-2}$
And by $v^{2}-u^{2}=2 a s$, we have
$v_{2}^{2}-v_{1}^{2}=2\left(a_{2}\right) s_{2} $
$\Rightarrow \,s_{2}=\frac{50 \times 50}{2 \times 10}=125\, m$
So, average speed for interval, $t=0$ to $t=15\, s$,
We have
$v_{\text {avg }}=\frac{s_{1}+s_{2}}{t_{1}+t_{2}}=\frac{250+125}{10+5}=\frac{375}{15}$
$=25 \,ms ^{-1}$