case 1 : time interval 0 to 2s
acceleration, a = 5m/s²
we know, acceleration is the rate of change of velocity with respect to time.
so, a = dv/dt = 5 ⇒v = 5t
so, velocity time graph is straight line.
at t = 2s , v = 5(2) = 10m/s
so, displacement = area under velocity-time graph
= 1/2 × 10m/s × 2s = 10m
case 2 : time interval 2s to 6s
acceleration = 0.
so, v = constant = 10m/s [ see figure]
so, displacement = area under velocity-time graph = area of rectangular part = 10m/s × (6s - 2s) = 40m
case 3 : time interval 6s to 8s
acceleration = -5m/s²
so, velocity, v = - 5t
so, displacement =1/2 × -5(8s-6s) × (8s - 6s) = -10m
now, net displacement = 10m + 40m - 10m = 40m
so, option (2) is correct choice.
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Answers & Comments
case 1 : time interval 0 to 2s
acceleration, a = 5m/s²
we know, acceleration is the rate of change of velocity with respect to time.
so, a = dv/dt = 5 ⇒v = 5t
so, velocity time graph is straight line.
at t = 2s , v = 5(2) = 10m/s
so, displacement = area under velocity-time graph
= 1/2 × 10m/s × 2s = 10m
case 2 : time interval 2s to 6s
acceleration = 0.
so, v = constant = 10m/s [ see figure]
so, displacement = area under velocity-time graph = area of rectangular part = 10m/s × (6s - 2s) = 40m
case 3 : time interval 6s to 8s
acceleration = -5m/s²
so, velocity, v = - 5t
so, displacement =1/2 × -5(8s-6s) × (8s - 6s) = -10m
now, net displacement = 10m + 40m - 10m = 40m
so, option (2) is correct choice.