Question

The table shows the distance covered in successive seconds by a body accelerated uniformly from rest
Time interval l II ||| |V
Distance261016
What is the speed of the body at the end of 4th second? (x is in meter)
a)4m * s ^ - 1;
b)1m * s ^ - 1;
c)14m * s ^ - 1;
d)16m * s ^ - 1​

Answer 1

Answer:

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.