Question

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

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Solution

## Step 1: GivenSpeed of the car, $u=126km/hr=35m/s$Final speed of the car, $v=0$Distance, $s=200m$Step 2: Determine the acceleration $a$Using the third equation of motion,${v}^{2}={u}^{2}+2as\phantom{\rule{0ex}{0ex}}{0}^{2}={35}^{2}+2\left(a\right)\left(200\right)\phantom{\rule{0ex}{0ex}}a=-\frac{{35}^{2}}{400}\phantom{\rule{0ex}{0ex}}=-3.065m/{s}^{2}$Negative acceleration suggests retardation.Step 3: Determine the time takenUsing the first equation of motion,$v=u+at\phantom{\rule{0ex}{0ex}}0=35+\left(-3.065\right)t\phantom{\rule{0ex}{0ex}}t=11.4s$Therefore, The retardation of the car is $-3.065m/{s}^{2}$ and it takes the car $11.4s$ to stop.

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