An object is moving with constant acceleration it's velocity is 48m per sec at the end of 10 second and becomes 68m per sec at the end of15 sec what would be the distance traveled by object in 15 second.
An object is moving with constant acceleration it's velocity is 48m per sec at the end of 10 second and becomes 68m per sec at the end of15 sec what would be the distance traveled by object in 15 second.
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.
hope it helps
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.
hope it helps
