Question

A car moving with constant acceleration covered the distance between two points $60.0m$ apart in $6.00s$. Its speed as it passes the second point was $15.0m/s$. At what prior distance from the first was the car at rest?

• A. $7.5m$
  
• B. $15m$
  
• C. $60m$
  
• D. $30m$
  

Answer 1

The correct option is A $7.5m$

From kinematics equation $S=ut+\frac{1}{2}g{t}^2;$ where $S=60m$; $t=6s$
$60=u\times 6+\frac{1}{2}a\times 6^2$...(i)
and the velocity at second point is given as $v=15m/s$
From kinematics equation $v=u+at$
$15=u+a\times 6.$...(ii)
solving equation (i) and (ii)
$a=\frac{5}{3}m/s^2;u=5m/s^2$
The car starts from rest with uniform acceleration
Using kinematics equation
$v^2=u^2+2aS$; $v=5m/s;u=0m/s;a=\frac{5}{3}m/s^2$
$\Rightarrow 25=0+2\times \frac{5}{3}\times S$
$S=7.5m$