Question

A car moving with uniform acceleration covers 450m in a 5 second interval and covers 700m in the next 5 second interval. The acceleration of the car is * 25m/s2 7m/s2 50m/s2 10m/s2

Answer 1

Given :

Initial distance of the car = 450 m .

Initial time taken = 5 s .

Initial velocity of the car ( u ) = 450 m / 5 s ⇒ 90 m/s .

Final distance covered by the car = 700 m .

Final time taken by the car = 5 s .

Final velocity of the car ( v )= 700 m / 5 s ⇒ 140 m/s .

Time taken to change the velocity ( t ) = 5 s .

Acceleration is the rate of change of velocity per unit time .

Change of velocity is = final velocity - initial velocity .

$a=\dfrac{v-u}{t}\\\\\implies a=\dfrac{140m/s-90m/s}{5s}\\\\\implies a=\dfrac{50m/s}{5s}\\\\\implies a=10m/s^2$

The acceleration of the car is 10 m/s².

NOTE :

The acceleration can be calculated by subtracting the final velocity and the initial velocity and by dividing it " with the time taken to change the initial and final velocity "

Answer 2

Solution:-

Given :-

For First 5 Seconds,

Initial Velocity = Total Distance Travelled/ Time Taken

=> Initial Velocity = (450/5) m/s

=> Initial Velocity = 90 m/s.

For Next 5 Seconds,

Final Velocity = Distance Travelled / Time Taken

=> Final Velocity = 700/5 m/s

=> Final Velocity = 140 m/s.

Now,

Change in Velocity due to Acceleration;

Change in Velocity= (Final Velocity- Initial Velocity)/ Time Taken

=> Change in Velocity = (140 - 90)/5

=> Change in Velocity = 50/5

=> Change in Velocity = 10 m/s^2.

Hence,

The Acceleration of the Car is 10 m/s^2.