Question

A body accelerate uniformly from rest at 0.2m/s for one-fifth of a minute. Calculate the distance covered by the body.​

Answer 1

Answer:

14.4 m

The distance covered by the body is 14.4 m.

Solution,

Initial velocity(u)= 0 m/s( according to Question, the body starts from rest.that's why initial velocity(u) becomes zero.

Acceleration (a)= 0.2 m/s

Time (t):

$\frac{1}{5} minute = \frac{1}{5 } \times 60 \: seconds$

$= 12 \: seconds$

Now,

Applying third equations of motion:

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 0 \times 12 + \frac{1}{2} \times 0.2 \times {(12)}^{2} \\s = 0 + 14.4 \\ s = 14.4 \: metre$

Thus, the distance covered by the body is 14.4 metre.

Further more information:

Application of equation of motion in different situation:

• When a certain object comes in motion from rest, in the case, initial velocity(u)= 0 m/s
• When a moving object comes in rest,in the case, final velocity(v)= 0 m/s
• If the object is moving with uniform velocity, in the case, (u=v)
• If any object is thrown vertically upward, in the case, acceleration (a)= -g
• When an object is falling from certain height, in the case, u= 0 m/s
• When an object is thrown vertically upwards in the case, final velocity at maximum height becomes 0.