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Question

# A CAR IS TRAVELLING AT A SPEED OF 10 ms. IT ACCELERAGS AT A RATE OF 1 ms FOR 40 SECS. THEN IT MOVES AT A CONSTANT SPEED FOR NEXT 2 MIN. FINALLY THE CAR COMES TO A STOP IN NEXT 90 SECS WITH A CONSTANT RETARDATION. WHAT IS THE TOTAL DISTANCE TRAVELLED BY THE CAR ?

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Solution

## $\mathrm{Case}\mathrm{I}:\phantom{\rule{0ex}{0ex}}\mathrm{Final}\mathrm{velocity}:\phantom{\rule{0ex}{0ex}}\mathrm{v}=\mathrm{u}+\mathrm{at}\phantom{\rule{0ex}{0ex}}=10+\left(1×40\right)\phantom{\rule{0ex}{0ex}}=50\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Distance}:\phantom{\rule{0ex}{0ex}}{\mathrm{s}}_{1}=\left(\frac{\mathrm{u}+\mathrm{v}}{2}\right)\mathrm{t}\phantom{\rule{0ex}{0ex}}=\left(\frac{10+50}{2}\right)40\phantom{\rule{0ex}{0ex}}=1200\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Case}\mathrm{II}:\phantom{\rule{0ex}{0ex}}\mathrm{Distance}:\phantom{\rule{0ex}{0ex}}{\mathrm{s}}_{2}=\mathrm{vt}\phantom{\rule{0ex}{0ex}}=50×120\left(2\mathrm{min}=120\mathrm{s}\right)\phantom{\rule{0ex}{0ex}}=6000\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Case}\mathrm{III}:\phantom{\rule{0ex}{0ex}}\mathrm{Distance}:\phantom{\rule{0ex}{0ex}}{\mathrm{s}}_{3}=\left(\frac{\mathrm{v}+\mathrm{v}\text{'}}{2}\right)\mathrm{t}\phantom{\rule{0ex}{0ex}}=\left(\frac{50+0}{2}\right)90\phantom{\rule{0ex}{0ex}}=2250\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Total}\mathrm{distance}:\phantom{\rule{0ex}{0ex}}\mathrm{d}=1200+6000+2250\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{9450}\mathbf{}\mathbf{m}$

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