A car starts from rest and move with constant acceleration of $4\frac{m}{s^2}$ and at the same time a motorcyclist moving with a constant speed of $36\frac{km}{hr}$ overtakes and passes the car. Find
a) How far beyond the starting point will the car overtakes the motorcyclist?
b) What will be the speed of the car at the time when it overtakes the motorcycle?
I am confused by part a of the question. Can anyone tell me, if the motorcyclist had a constant speed of $36\frac{\mathrm{km}}{\mathrm{hr}}$ and the car had a constant acceleration or $4\frac{\mathrm{m}}{\mathrm{s}^2}$, how the bike overtook and passed by the car so quickly? Also it is not clear here from where the bike started?
Best Answer
Let's say car and bike be at rest at $1pm$ so, $v_c=0$ and $v_b=0$.
Calculations for motion of car:
Since car is moving with constant acceleration,
At 1:00:00pm, $v_c=0m/s$, $S_c=0m$
At 1:00:01pm, $v_c=4m/s$, $S_c=4m$
At 1:00:02pm, $v_c=8m/s$, $S_c=12m$
At 1:00:03pm, $v_c=12m/s$, $S_c=24m$
At 1:00:04pm, $v_c=16m/s$, $S_c=40m$
Calculations for motion of bike:
Since bike is moving with constant speed,
At 1:00:00pm, $v_b=0$, $S_b=0m$
At 1:00:01pm, $v_b=10m/s$, $S_b=10m$
At 1:00:02pm, $v_b=10m/s$, $S_b=20m$
At 1:00:03pm, $v_b=10m/s$, $S_b=30m$
At 1:00:04pm, $v_b=10m/s$, $S_b=40m$
So from above calculations that in $4$ $seconds$ your car will overtake the bike after covering the distance of $40$ $meters$.