Question

a car accelerates uniformaly by 1m/S2 from 5m/second to 10m/second calculate the distance covered by car in that time (explain) ​

Answer 1

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

Answer 2

Answer:

Given

1. acceleration = 1m/s^2
2. initial velocity u = 5mls
3. final velocity. v = 10 m/s

By using third equation if motion

v^2 - u^2 = 2as ----( i )

putting values in the equation ( i )

(10)^2 -(5)^2 = 2(1)s

100 - 25 = 2 s

75 /2 = s

37.5 m = s

so, distance covered by car is 37.5m