Question

A car starts from rest and moves along the x-axis with constant acceleration 5ms−2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?

pls tell​

Answer 1

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

Answer 2

GIVEN

A car starts from rest and moves along the x-axis with constant acceleration 5 m/s² for 8 seconds.

◙ Initial velocity, u = 0 m/s

◙ Acceleration, a = 5 m/s²

◙ Time, t = 8 secs

◙ Velocity is constant.

$\underline{\rule{200}2}$

TO FIND

The distance covered by the car in 12 s since it started from rest, if its velocity remains constant.

$\underline{\rule{200}2}$

SOLUTION

Using the first law of Kinematics :-

$\boxed{\color{fuchsia} {\mathbf{v=u+at}}}$

Substituting the values :-

$\sf{\longmapsto v=0+5(8)}\\\\\boxed{\sf{\longrightarrow v=40\ m/s}}$

$\rule{120}{1.6}$

For the distance covered in 12 secs since the starting, using the second law of kinematics :-

$\boxed{\color{fuchsia} {\mathbf{s=ut+\frac{1}{2} at^2}}}$

Substituting the values :-

$\sf{\longmapsto s=0(12)+\dfrac{1}{2} (5)(12)^2}\\\\\sf{\longmapsto s=\dfrac{144 \times 5}{2} }\\\\\sf{\longmapsto s=72\times5}\\\\\large\boxed{\boxed{\color{green} {\sf{\longmapsto s=360\ m}}}}$

Therefore, the car covers 360 m in 12 seconds since it started from the rest.