Question

(12) A car starting from rest acquires a speed of 40m/s in a distance of 80m. Calculate
: (i) the time in which this speed is acquired (ii) the acceleration and (iii) the
speed of the car at the end of 3 seconds. ​

Answer 1

Answer:

(i) 2 s

(ii) 20 m/s^2

(iii) 28.66 m/s

Explanation:

(i)

Time= distance/speed

= 80/40

= 2 s

(ii)

acceleration= velocity/time

= 40/2

= 20 m/s^2

(iii) given

time= 3 s

speed = distance / time

= 80/3

= 26.66 m/s

Answer 2

Given:-

→ Initial speed of the car = 0

→ Final speed of the car = 40m/s

→ Distance covered = 80m

To find:-

→ Time in which, final speed is acquired

→ Acceleration of the car

→ Speed of the car at the end of 3s

Solution:-

Firstly, let's calculate the acceleration of the car by using the 3rd equation of motion :-

=> v² - u² = 2as

=> (40)² - 0 = 2×a×80

=> 1600 = 160a

=> a = 1600/160

=> a = 10m/s²

Now, let's calculate the time taken by the car to acquire the final speed by using the 1st equation of motion :-

=> v = u + at

=> 40 = 0 + 10t

=> 40 = 10t

=> t = 40/10

=> t = 4s

Let the velocity of the car at the end of 3 seconds be v' . Then by 1st equation of motion :-

=> v' = u + at

=> v' = 0 + 10(3)

=> v' = 0 + 30

=> v' = 30m/s

Thus :-

• Time taken by the car to acquire the

final speed is 4 seconds .

• Acceleration of the car is 10m/s² .

• Speed of the car at the end if 3 seconds

is 30m/s.