Question

Solution of question no. 3?

Answer 1

a = 3 - 4t
v = 3t - 2t² + 2
x = 1.5t² - 2t³/3 + 2t

v = 3t - 2t² + 2
If v = 0
2t² - 3t - 2 = 0
Solve this above polynomial equation.
We will get t = 2 and t = -0.5
Neglect -ve term as our starting point is t = 0

Displacement in t = 2 seconds
x = 1.5t² - 2t³/3 + 2t
= (1.5 × 2²) - (2 × 2³/3) + (2 × 2)
= 10 - 16/3
= 14/3 m

Displacement in t = 3 seconds
x = 1.5t² - 2t³/3 + 2t
= (1.5 × 3²) - (2 × 3³/3) + (2 × 3)
= 3/2 m

Understand this:
Particle starts from t = 0 and stops at t = 2 s. Then it reversed its direction and travelled Δx distance for next 1 s.

Displacement in 3 s is 3/2 m
Displacement in 2 s is 14/3 m
Δx = 14/3 - 3/2
= 19/6 m

DISTANCE travelled in 3 seconds
= Distance travelled in 2 seconds + Δx
= 14/3 + 19/6
= 47/6 m

(D) is the correct option.