A car accelerates from rest at a constant rate of 2m/s2 for some time. Then it retards at a constant rate of 4m/s2 and comes to rest. It remains in motion for 6 see.

Maximum speedvmax=αβtα+β α and β are the acceleration and retardation of the bodyvmax=2×46×6=8 m/stotal distance travelleds=12αβα+βt2s=122×4636=24 m

A car accelerates from rest at a constant rate of 2m/s2 for some time. Then it retards at a constant rate of 4m/s2 and comes to rest. It remains in motion for 6 see.

Maximum speedvmax=αβtα+β α and β are the acceleration and retardation of the bodyvmax=2×46×6=8 m/stotal distance travelleds=12αβα+βt2s=122×4636=24 m

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A car accelerates from rest at a constant rate of 2m/s² for some time. Then it retards at a constant rate of 4m/s²and comes to rest. It remains in motion for 6 see.

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It travelled a total distance of 24m

It travelled a total distance of 18m

Its maximum speed is 6m/s

Its maximum speed is 8m/s

answer is A, C.

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Detailed Solution

Maximum speed

$v_{\max} = \frac{αβt}{α + β} α and β are the acceleration and retardation of the body$

$v_{\max} = \frac{2 \times 4}{6} \times 6 = 8 m / s$

total distance travelled

$s = \frac{1}{2} (\frac{αβ}{α + β}) t^{2}$

$s = \frac{1}{2} \frac{2 \times 4}{6} 36 = 24 m$

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A car accelerates from rest at a constant rate of 2m/s2 for some time. Then it retards at a constant rate of 4m/s2 and comes to rest. It remains in motion for 6 see.

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A car accelerates from rest at a constant rate of 2m/s² for some time. Then it retards at a constant rate of 4m/s²and comes to rest. It remains in motion for 6 see.