An object having a velocity 4.0 $\frac{m}{s}$ is accelerated at the rate of 1.2 $\frac{m}{s^{2}}$ for 5.0 s. Find the distance travelled during the period of acceleration.

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Solution

Distanc e travelled, S =ut+ $\frac{1}{2} a t^{2}$
= $4.0 \times 5 + \frac{1}{2} \times 1.2 \times 5^{2}$
= $20.0 + \frac{1}{2} \times 12 \times 25$
=35 m.

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An object having a velocity 4.0 ms is accelerated at the rate of 1.2 ms2 for 5.0 s. Find the distance travelled during the period of acceleration.

Distanc e travelled, S =ut+12at2 = 4.0×5+12×1.2×52 = 20.0+12×12×25 =35 m.

An object having a velocity 4.0 $\frac{m}{s}$ is accelerated at the rate of 1.2 $\frac{m}{s^{2}}$ for 5.0 s. Find the distance travelled during the period of acceleration.

Distanc e travelled, S =ut+ $\frac{1}{2} a t^{2}$
= $4.0 \times 5 + \frac{1}{2} \times 1.2 \times 5^{2}$
= $20.0 + \frac{1}{2} \times 12 \times 25$
=35 m.