Question

a car starts from rest and moves along the x axis with constant acceleration 5m/s Square for 8 seconds. if it then continues with with constant velocity what distance will it cover in 12s since it started from rest??​

Answer 1

a=5m/s^2

u=0m/s

NOW

t=12 sec

s=ut+1/2at^2

s=0×8+1/2×5×8×8

=0+320

=320m

THANK YOU

PLZ MARK ME BRAINLIST

Answer 2

ANSWER:-

320 m

Explanation :-

initial velocity(U) = 0

Final velocity(V) = ?

Acceleration(a) = 5m/s²

time = 8 sec

{ first find the velocity qt the end of 8 sec then after this it will become constant}

•°• V = u + at

V= 0 + 5 × 8

V= 40m/s{ constant velocity}

Velocity at the end of 8Th second is 40m/s

The distance travelled in 8 sec

is by using the formula :-

V² = U²-2aS

(40)² = 0²- 2 × 5 × s

S = 1600/10

S = 160m

The distance travelled during the time 8 sec is 160 m

Now,

During the last 4 sec the velocity 40m/s is constant

Initial velocity (U)= 40/s

Final velocity(V) = 40 m/s{ °•°at the end of 12 sec the final velocity is 40 m/s}

By using the formula

S = v+u/2 × t [ v+u/2 = average velocity]

S = 40 + 40/2 × 4

S = 160 m

Now,

Total distance travelled is =

{160+160}m

=> 320 m