Question

A car starts from rest and moves along the x axis with constant acceleration 5 m/s2 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?

Answer 1

given, u = 0 m/s (rest)
a = 5m/s^2
time t = 8 sec
velocity of car after 8 sec,
v= u +at
v = 0 +( 5×8)=40m/s
also,
v^2 = u^2 + 2as
=>40^2 = 0+(2×5×s)
=> s = 1600/10 =160m
i.e distance covered in first 8 sec = 160m

now,
it further moves with const. velocity of 40m/s
then,
distance covered by the car in 4 sec ( 8th sec to 12th sec) = 40×4 = 160m
hence, the total distance covered by the car in 12sec from starting = 160 +160 = 320m
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I hope this will help you..

Answer 2

Answer:

320m after 12 sec

Explanation:

speed after 8 sec

v=u+at

=5×8

=40m/s

Now

distance covered in 8 s

s=ut+1/2at^2

=1/2 × 5 × 64

=160m

Distance covered by constant speed=40×4=160m

total distance=160+160=320m

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