Question

a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance travelled in 5s​

Answer 1

The velocity acquired in 5 s is 10 m/s and the distance travelled in 5 s is 25 m.

Explanation:

As acceleration is the measure of change in velocity with respect to time, if the acceleration and time is known , we can easily determine the velocity using the first equation of motion.

v = u +at

Here v is the final velocity which we have to determine, u is the initial velocity which is given as zero and acceleration is given as 2 ms⁻².

So, velocity of the body at 5 s will be v = 0 + (2× 5 s), v = 10 m/s.

As the given problem stated that motion is occurring at constant acceleration, then the distance travelled by the body can be obtained from second equation of motion.

s = ut + 0.5 at²

So, here the displacement s will be equal to the distance as the starting position is considered as zero. And initial velocity u is also zero, a is given as 2 ms⁻² and the time is 5s.

Then, s = (0×5)+(0.5×2×5×5)=25 m

So, the velocity acquired in 5 s is 10 m/s and the distance travelled in 5 s is 25 m.