Compute the displacement of the particle:
The acceleration time diagram for particle is as specified below .Computed velocity of the particle at several time intervals, plot velocity time diagram and computed the displacement of the particle at several intervals. Suppose initial velocity v_{0 }= 0 and initial displacement s_{0} = 5.30 cm.
Solution
We have
dv/ dt = a, or dv = a dt
∴ v_{1} - v_{0} = Area of (a - t ) diagram between t = 0 and t =1 seconds
∴ v_{1} = v_{0 } + 10 × 1 v_{0} = 0
= 10 cm / sec
Likewise,
∴ v_{2} - v_{0} = Area of (a - t ) diagram between t = 0 and t = 2 seconds
= v_{0} + 10 × 2
= 20 cm / sec
And v_{3 } = 30 cm / sec
We have ds/ dt = v, or ds = v dt
or s_{1} - s_{0 } = Area of (v - t ) diagram among t = 0 and t =1 seconds
∴ s1 = s0 + (½) × 1 × 10
= 5.30 + 5.0
= 10.30 cm .
s_{2} - s_{0} = Area of (v - t ) diagram among t = 0 and t = 2 seconds
= 5.3 + (½) × 20 × 2
= 25.3 cm .
s_{3} - s_{0} = Area of (v - t ) diagram among t = 0 and t = 3 seconds
s_{3 } = 5.3 + (½) × 30 × 3 = 50.3 cm .