A car is accelerating uniformly as it passes two check points that are 30 m apart. The time taken between check points is 4.0 s and the car’s speed at the first check point is $5.0 m s^{- 1}$ . Then,

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the acceleration of car is $2.5 m s^{- 2}$

the velocity of car at second check point is $10 m s^{- 1}$

the acceleration of car is $2.25 m s^{- 2}$

the velocity of car at second check point is $20 m s^{- 1}$

answer is A.

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Detailed Solution

$S = (\frac{U + V}{2}) t$

$V = \frac{2 S}{t} - U = \frac{60}{4} - 5 = 10 m / s$

$a = \frac{V - U}{t}$

$a = \frac{10 - 5}{4} = 1.25 m / s^{2}$

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