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A certain force exerted for $1.2 s$ raises the speed of an object from $1.8 m / s$ to $4.2 m / s$ . Later the same force is applied for $2 s$ . How much will the speed change in $2 s$ ?

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2.5 m / s

4 m / s

5 m / s

3.5 m / s

answer is B.

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Detailed Solution

A certain force exerted for

1.2 s

raises the speed of an object from

1.8 m / s

4.2 m / s

. Later the same force is applied for

2 s

. The change of speed in

2 s

4 m / s

.
Given data in the problem,
Initial velocity,

u = 1.8 m / s

Final velocity,

v = 4.2 m / s

Time duration for force applied,

t = 1.2 s

By first law of motion,

v = u + at

Where,

v = final velocity

u = initial velocity

a = acceleration

t = time

And by rearranging the equation, the acceleration of a body is given as

\Rightarrow a = \frac{(v - u)}{t}

\Rightarrow a = \frac{4.2 - 1.8}{1.2}

\Rightarrow a = \frac{2.4}{1.2}

\Rightarrow a = 2 m s^{- 2}

and change of speed is given by
again, the same force applied for time,

t = 2 s

change of speed = a \times t

= 2 \times 2

= 4 m / s

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