Question

A car starts from rest and most
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

Answer:

Explanation:

In the above Question , the following information is given -

A car starts from rest and moves along the x-axis with constant acceleration 5ms−2 for 8 seconds.

it then continues with constant velocity.

We have to find the distance covered by the car in 12 seconds after it started from rest .

Solution -

From various laws of motion , we know that -

Acceleration = ∆ V / t

=> ( v - u ) / t

Here , the car started from rest .

So ,

u = 0

It travelled for 8 seconds .

Hence , t = 8 seconds .

We also have the value of acceleration of the car given as 5 m / s^2

So ,

5 = ∆v / 8

=> 5 = v / 8 [ as u = 0 ]

=> v = 40 m / s.

Now ,

Distance travelled by the car in these 8 seconds

=> S = ut + ( 1 / 2 ) at ^ 2

=> S = ( 1 / 2 ) × 5 × 64

=> 5 × 32

=> 160 m.

Now , it travels with a constant velocity of 40m/s for next 4 seconds .

So, distance travelled in next 4 seconds = 160 m.

Hence total distance travelled by the car in 12 seconds -

=> 160 m + 160m

=> 320 m .

Answer 2

Answer: 320m

s1=ut+(1/2)at^2

=0+  1/2×8²

=160 m

the velocity attained by 8 seconds is given by

v=u+at

  =0+5×8=40 m/s

the distance travelled in next 4 seconds is

s2=40×4=160 m

So, the total distance traveled by the car is

s1+s2=160+160=320 m

hope it helps!!!!!