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Standard IX

Physics

Derivation of Velocity-Time Relation by Graphical Method

A car travels...

Question

A car travels a distance 100 m with a constant acceleration and average velocity of $20 m s^{- 1}$ . The final velocity acquired by the car is $25 m s^{- 1}$ . Find the acceleration of the car.

1 m s^{- 2}

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2 m s^{- 2}

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0 m s^{- 2}

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4 m s^{- 2}

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Solution

The correct option is B $2 m s^{- 2}$
$Average velocity = \frac{v + u}{2}$
$20 = \frac{25 + u}{2}$
$u = 15 m / s$

$v = u + a t$
$25 = 15 + a \times 5$
We get, $a = 2 m / s^{2}$

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A car travels a distance 100 m with a constant acceleration and average velocity of 20ms−1. The final velocity acquired by the car is 25ms−1. Find the acceleration of the car.

The correct option is B 2ms−2Average velocity=v+u220=25+u2u=15 m/sv=u+at25=15+a×5We get, a=2 m/s2

A car travels a distance 100 m with a constant acceleration and average velocity of $20 m s^{- 1}$ . The final velocity acquired by the car is $25 m s^{- 1}$ . Find the acceleration of the car.

The correct option is B $2 m s^{- 2}$
$Average velocity = \frac{v + u}{2}$
$20 = \frac{25 + u}{2}$
$u = 15 m / s$

$v = u + a t$
$25 = 15 + a \times 5$
We get, $a = 2 m / s^{2}$