Question

[2]
A body starts from rest and accelerates uniformly with 8 ms^-2
?. Using the method of integration
calculate the distance covered by the body in 5 to 8 second.​

Answer 1

Given :-

Acceleration of body = 8 ms-²

Here, we will integrate the acceleration with respect to time i.e. 't'. Also we need to double integrate to find the distance travelled.

At first integration we we will find Velocity of body.

$\int{dv} = \int{a\:dt}$

$v = \int{8\:dt}$

v = 8t

Now again integrating the function w.r.t 't'

$\int{dS} =\int{v\:dt}$

$S = \int^{8}_{5}{8t\:dt}$

$S = [4{t}^{2}]^{8}_{5}$

S = [4(8)²] - [4(5)²]

S = [256 - 100] m

S = 156 m

Hence,

Distance travelled = S = 156 m