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Standard XII

Physics

Displacement

A car acceler...

Question

A car accelerates from rest at $5 {m / s}^{2}$ and then retards to rest at $3 {m / s}^{2}$ . The maximum velocity of the car is $30$ $m / s$ . The distance covered by the car is

150 m

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240 m

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300 m

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360 m

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Solution

The correct option is A 240 m
Given maximum Velocity is 30 m/s
Distance travelled during acceleration is given by
$30^{2} - 0 = 2 \times 5 \times S_{1}$
Hence , $S_{1} = 90 m$
Distance travelled during retardation
$0 - 30^{2} = 2 \times - 3 \times S_{2}$
Hence , $S_{2} = 150 m$
$∴$ Total distance travelled is $90 + 150 = 240 m$

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A car accelerates from rest at 5m/s2 and then retards to rest at 3m/s2. The maximum velocity of the car is 30 m/s. The distance covered by the car is

The correct option is A 240 mGiven maximum Velocity is 30 m/s Distance travelled during acceleration is given by 302−0=2×5×S1Hence , S1=90mDistance travelled during retardation 0−302=2×−3×S2Hence , S2=150m∴ Total distance travelled is 90+150=240m

A car accelerates from rest at $5 {m / s}^{2}$ and then retards to rest at $3 {m / s}^{2}$ . The maximum velocity of the car is $30$ $m / s$ . The distance covered by the car is