Question

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

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Solution

## Step 1. Given data:$Initialvelocity\left(u\right)=126Km{h}^{-1}\phantom{\rule{0ex}{0ex}}Initialvelocity\left(u\right)=\frac{126}{3600}×1000\phantom{\rule{0ex}{0ex}}Initialvelocity\left(u\right)=35m{s}^{-1}$$Dis\mathrm{tan}ce\left(s\right)=200m$$Finalvelocity\left(v\right)=0$, since the car will stop after some timeStep 2. Formula used:3rd Equation of Motion${v}^{2}={u}^{2}+2as$Where $v$ is the final velocity,$u$ is initial velocity,$a$ is acceleration and$s$ is distanceFirst Equation of Motion$v=u+at$Where $t$ is the time takenStep 3. Calculating retardation of the car,${v}^{2}={u}^{2}+2as$$a=-\frac{{u}^{2}}{2s}$ $a=\frac{-{u}^{2}}{2s}$ $=-3.06\frac{m}{{s}^{2}}$The negative sign indicates that the car is retarding. Retardation is nothing but negative acceleration.Step 4. Calculating time taken,$v=u+at$$t=\frac{-u}{a}$$=\frac{-35}{-3.06}$$t=11.43s$Hence, the retardation of car is $-3.06m{s}^{-2}$ and time taken by the car to stop is $t=11.43s$

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