Question

A car travelling at a speed of 80 kmph is brought to a speed of 60 kmph within a distance of 30 m by applying brakes. On travelling further at a speed of 60 kmph, the driver of the car saw an overturned truck across and he applied the brakes again. The car stopped just close to the overturned vehicle. What was the distance (in kmph) between the overturned truck and the car when the driver first saw the overturned truck? The reaction time of the driver may be taken as 2 seconds.

• A. 71.86

Answer 1

The correct option is A 71.86
Initial speed of Car, U = 80 kmph
$=80\times \frac{5}{18}m/s$
$=22.22m/s$

Final speed of car, V = 60 kmph
$=60\times \frac{5}{18}m/s$
$=16.66m/s$

Distance travelled , S = 30 m
$\Rightarrow v^2=u^2+2aS$
$\Rightarrow 277.55=493.72+2\times a\times 30$
$a=-3.60m/s^2$

$\text{Retardation is}3.60m/s^2$

We know that,
$a=fg$
$\Rightarrow 3.60=f\times 9.81$
$\Rightarrow f=0.367$

Now, stopping sight distance.

$\text{SSD}=vt+\frac{v^2}{2gf}$

Where
$v=16.66m/s$
$\text{reaction time,}t=2sec$
$f=0.367$

$\Rightarrow \text{SSD}=(16.66\times 2)+\frac{(16.66{)}^2}{2\times 9.81\times 0.367}$
$=33.32+38.54$
$=71.86m$

The distance between an overturned truck and car is 71.86 m so that the car stops just close to the overturned vehicle.