Question

A car starts from rest and moves along the x-axis with constant acceleration 5 m/s for 8 seconds.
If it then continues with constant velocity, what distance will the car cover in 12 seconds since
it started from the rest.










pls answer​

Answer 1

since the car started from rest so u = 0m/s

a = 5m/s t=8s constant v net time = 12 secs

so here in this case when there is a constant accelaration ...our distance covered will be

s= ut + 1/2at ^ 2

s= 0 + 160 s = 160 m in first case

when the car is in constant velocity the in that case we got

v = u + at

v = 0 + 60 hence v = 60m/s therefore distance in second case is vt that is 300 m

hence our net distance will be 300 + 160m

= 460 m answer

please mark as brainalist rofl

Answer 2

Answer:

The answer to your question is 320m.

Explanation:

Given that

Displacement in first 8 seconds.

u=0

a= 5m/s^2

t= 8 seconds

v= u+at

v= 0+5(8)

=40m/s

using v^2- u^2=2aS

S= 160 m.

Displacement in last 4 seconds.

u= 40m/s ( constant velocity as calculated earlier)

a=0

t= 4 seconds.

Using S=ut+1/2at^2

as a=0.

so, S=ut

S=40(4)

=160 m

Therefore the total displacement is 160+160 = 320m