Question

A car moving at $30\, m/s$ slows uniformly to a speed of $10\, m/s$ in a time of $5\, s$. Determine the distance moved in the third second.

• A. 10 m
• B. 20 m
• C. 40 m
• D. 5 m

Answer 1

Answer: (B)

Rate of change of speed gives acceleration.
$\therefore $ Acceleration/Retardation
$=\frac{\text { change in speed }}{\text { time }}$
$\therefore $ Retardation $(a)=\frac{30-10}{5}=-4\, m / s ^{2}$
From equation of motion:
when $t=2\, s,\, s_{1}=u t+\frac{1}{2} a t^{2}$
$=(30 \times 2)-\frac{1}{2} \times 4 \times(2)^{2}=52\, m$
When $t =3\, s,\, s_{2}=30 \times 3-\frac{1}{2} \times 4(3)^{2}=72\, cm$
$\therefore s_{2}-s_{1}=72-52=20\, m$