Question

E-6. starting glfrom rest at t = 0, a car moves in a straight line with an acceleration given by the accompanying
graph. The speed of the car at t = 3 sis :
a(ms)
ONWAO
(A) 1 ms-1
1 2 3 4 5 6 t(sec)
(B) 2 ms-1
(C) 6.0 ms-1
(D) 10.5 ms-1​

Answer 1

Answer:

Change in velocity = area under acceleration time graph

Velocity of car at t=0s is 0.

Lets say velocity of car at t=3s is V

Change in velocity from t=0 to t=3s = V - 0.

Area under acceleration graph from t=0 to t=3s = area of shaded region = area of triangle OAB - area of triangle CDB = (0.5*5*5) - (0.5*2*2) = 12.5-2=10.5 m/s

So, V - 0 = 10.5 m/s

V = 10.5 m/s

Answer 2

Answer:Change in velocity = area under acceleration time graph

Velocity of car at t=0s is 0.

Lets say velocity of car at t=3s is V

Change in velocity from t=0 to t=3s = V - 0.

Area under acceleration graph from t=0 to t=3s = area of shaded region = area of triangle OAB - area of triangle CDB = (0.5*5*5) - (0.5*2*2) = 12.5-2=10.5 m/s

So, V - 0 = 10.5 m/s

V = 10.5

Explanation: