Kinematics and integration maths question

The velocity (v m/s) at time t seconds (t ≥ 0) of a particle is given by v=2t^2-8t+6. It is initially 4 m to the right of a point of origin. Find:
a) its displacement and acceleration at any time
b) its displacement when the velocity is zero
c) its acceleration when the velocity is zero.

d= integral v = 2/3 t^3 -4t^2 + 6t +K
at t = 0, d = 4 ....so, K =4 hence
displacement is given by d = 2/3 t^3 -4t^2 + 6t +4


a = dv/dt = 4t - 8 is acceleration at any time.

for v=0, t^2 -4t +3 =0 ie ( t-3)(t-1) =0

velocity is zero at t=1, t=3

substitute in eqn. for d and a

d ( v=0) = 2/3 +6 +4-4 = 20/3 and d ( v=0) = 18- 36 + 18+4 = -22

a( v=0) = 1-8 = -7 and a ( v=0) =12-8 = 4

The person's answers for b) and c) are wrong. Has this guy ever taken physics? When velocity is zero, the displacement is a constant and in this case, the acceleration is zero since the velocity is zero.

Report Abuse

a) Int(ds) = Int(2t^2 - 8t + 6) with given condition that s(0) = 4
s(t) = (2/3)t^3 - 4t^2 + 6t + 4
a(t) = 4t - 8

b) When the velocity is zero this means that s(t) is a constant, so s(t) = 4
c) The acceleration would be 0.