Question

A body starts with an initial velocity of 10 ms-1 and acceleration 5ms-2 find distance covered bi it in 5s

Answer 1

Sorry friend I am compelled to answer the question from picture.

Answer 2

hiii!!!

here's ur answer...

given :-

u (initial velocity) = 10ms^-1

a (acceleration) = 5ms^2

t (time taken) = 5s

we can solve it by two methods.. first we can find it's final velocity and can use third equation (2as=v²-u²) or second equation (ut+1/2at²) without finding it's final velocity (v)

we know that a = v-u/t

==> 5ms^-2 = ( v - 10 ) / 5

==> 5 × 5 = v - 10

==> 25 = v - 10

==> 25 + 10 = v

==> 35 = v

hence, final velocity of the body is 35 ms^-1

by using third equation 2as = v² - u² (s=distance)

=> 2 × 5 × s = 35² - 10²

=> 10s = 1225 - 100

=> 10s = 1125

=> s = 1125/10

=> s = 112.5 m

hence, the distance covered by the body in 5secs is 112.5m

now, we will find the distance by using second equation s = ut+1/2at²

=> s = ( 10 × 5 ) + 1/2 × 5 × 5²

=> s = 50 + 5/2 × 25

=> s = 50 + 125/2

=> s = 50 + 62.5

=> s = 112.5 m

hope this helps..!!