Question

. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance travelled. [Ans. a = 11.2 m/s and s = 79.8 m]​

Answer 1

Answer:

Here, initial velocity is u=18.5m/s and final velocity is v=46.1m/s

Time taken t=2.47s

If a be the acceleration of the car.

Using v=u+at,

46.1=18.5+a(2.47)

⟹ a=11.17m/s

2

If S be the distance traveled by car.

Using formula v

2

−u

2

=2aS,

(46.1)

2

−(18.5)

2

=2(11.17)S

⟹ S=79.8m

Answer 2

Answer:

The acceleration of car is 11.2 m/s^2 and distance (travelled) is 79.8 m.

Explanation:

The both accelerations in first and second case are given as follows -  

Vi = 18.5 m/s and Vf = 46.1 m/s  

and time for both accelerations is 2.47 m/s  

a = v/t  

= 46.1 - 18.5 / 2.4

a = 11.2 m/s^2  

d = vi × t + 0.5 × a × t ^2  

d = 18.5 m/s × 2.47 + 0.5 × (11.2 m/s^2) × (2.47 s)^2  

d = 45.7 m + 34.1 m  

d = 79.8 m

Hope this helps...

Thanks  

Regards

Teevra Aryan Das