Question

A car moving at $30m/s$ slows uniformly to a speed of $10m/s$ in a time of $5s$. 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 $(a)=\frac{30-10}{5}=-4m/s^2$
From equation of motion:
when $t=2s,s_1=ut+\frac{1}{2}a{t}^2$
$=(30\times 2)-\frac{1}{2}\times 4\times (2{)}^2=52m$
When $t=3s,s_2=30\times 3-\frac{1}{2}\times 4(3{)}^2=72cm$
$\therefore s_2-s_1=72-52=20m$