An object is moving with constant acceleration. Its velocity is 48 mps at the end of 10s and becomes 68 mps at end of 15s. What will be distance covered by object in 15s?
Answers
Velocity of the body at the end of 10s = 48 m/s,
Velocity of the body at the end of 15s = 68 m/s
Change in velocity in 5 seconds= 68m/s - 48m/s= 20m/s,
Acceleration (taking it as uniform) = (change in velocity)/time interval in which the change in velocity occurs= (20 m/s)/5s= 4 m/s².
We need to find initial velocity. Using the rel4ation;
v = u + a t. We know that v= 48 m/s, a= 4 m/s², t= 10s. Substituting these values in v= u + a t, we get,
u = v - a t = 48 m/s - 4 m/s² ×10s = 48 m/s - 40 m/s = 8 m/s.
Using the relation s = u t + ½ a t², where u = 8 m/s; a= 4 m/s²; t = 15 s, we get,
s = 8 m/s×15s + ½ 4 m/s² × (15s)² = 120 m + 2×225m = 120m + 450m= 570m.
We could have got s from the relation v² - u² = 2 a s.
s = ( v² - u²)/2 a = (68² - 8²)/2×4 = (4624 - 64)/8= 4560/8= 570m.