Question

Solve the following : The position of a particle is given by the function s (t) = 2t² + 3t – 4. Find the time t = c in the interval 0 ≤ t ≤ 4 when the instantaneous velocity of the particle equal to its average velocity in this interval.

Answer 1

s (t) = 2t² + 3t – 4
∴ s(0) = 2(0)² + 3(0) – 4 = – 4
and
s(4) = 2(4)² + 3(4) – 4 = 32 + 12 – 4 = 40

∴ average velocity = `(s(4) - s(0))/(4 - 0)`

= `(40- (-4))/(4)`
= 11
Also , instantaneus velocity = `"ds"/dt`

= `d/dt(2t^2 + 3t - 4)`
= 2 x 2t + 3 x 1 – 0
= 4t + 3

∴ instantaneus velocity at t = c is `("ds"/dt)_(t = c) = 4c + 3`

When instantaneous velocity at t:c equal to its average velocity, we get
4x + 3  11
∴ 4c = 8
∴ c = 2 ∈[0, 4]
Hence, t = c = 2.