Question

Q4 A car accelerates uniformly by 1m/s 2 from 5 to 10m/s in 5 s. Calculate distance covered by the car in that time.​

Answer 1

Initial velocity of the car (u)= 5 m/s

Final velocity of the car (v) = 10 m/s

Time taken for change in velocity (t)= 5 s

Therefore acceleration (a)= [10 m/s - 5 m/s]/ 5 s = 5 m/s /5s = 1 m/s²

Distance covered during these 5 seconds, is found using the relation

s = u t + ½ a t² = 5 m/s× 5 s + ½ 1 m/s² × 5² s²=25 m+ 12.5 m = 37.5 m

So car covers 37.5 m during the period its velocity changes from 5 m/s to 10 m/s.

Answer 2

Answer:

$\huge \rm \red{Answer}$

According to Second equation of motion

(time-position relation) s = ut + (1/2)at2

acceleration (a) = 1 m/s2

Initial velocity of the car (u)= 5 m/s

Final velocity of the car (v) = 10 m/s

Time taken for change in velocity (t)= 5 d

Distance covered during these 5 seconds, is found using the relation s = u t + (1/2)at2

= 5 m/s X 5 s + (1/2)1 m/s2 X 52 s2

= 25 m+ 12.5 m = 37.5 m

• Hence distance covered by the car in that time is 37.5 m