Sol. Initial velocity

a t = 10 sec.

Area under a.t. curve b/w 0 & 10 sec.

Δ V = 2 × 10 = 20m/s

V f – V = 20 m/s

V f = 20 m/s

at t = 20 sec.

Total area under a-t curve between 0 & 20 sec.

Δ V = 2 × 10 – 3 × 5

Δ V = 20 – 15

V f ′ – V i = Δ V = 5m/s

V f ′ = 5m/s.

262 m

First method:

Calculation of displacement directly from a–t curve by using following formula.

Δ s = ut 0 + (area under a–t curve) (t 0 – t c )

u → initial velocity

t 0 → total time

t c → abscissa of centroid of corresponding area.

Δ s = ut 0 + (area under a–t cure) (t 0 –t c )

Δ s = 20 × 15 – 15 × 2.5

= 300 – 37.5

Δ s = 262.5m

Displacement & distance traveled are same in this case.

Second method:

Area under velocity-time curve gives displacement.

Δ s = × 10 × 20 + 20 × 5 + × [20 + 5] × 5

= 100 + 100 + 25 × 2.5

Δ s = 262.5 m