Question

A car moving along a straight highway with a speed of $126kmph$ is brought to a stop within a distance of $200m$. What is the acceleration of the car and how long does it take for the car to stop?

• A. $-3.06m{s}^{-2},11.43s$
• B. $-6.03m{s}^{-2},11.43s$
• C. $-3.06m{s}^{-2},10.34s$
• D. $-6.03m{s}^{-2},10.34s$

Answer 1

The correct option is A $-3.06m{s}^{-2},11.43s$

Initial velocity $u=126km/hr$ or $35m/s$
Final velocity $=0m/s$ (stopped)
Distance $s=200m$
Use newton's 3rd law of motion
$v^2=u^2+2as$ ; a is acceleration or -a if it is deceleration
$0={35}^2+2\times a\times 200$
$400a=-1225$
$a=-3.0625m/s^2$ ---------> This is retardation of the CAR to find time

Use first law of motion
$v=u+at$; first law of motion,
$0=35-3.0625t$
$t=35/3.0625=11.4285sec$