Question

The displacement of a particle at time t is given by s = 2t³ − 5t² + 4t − 3. Find the velocity and displacement at the time when the acceleration is 14 ft/sec².

The displacement of a particle at time t is given by s = 2t³ − 5t² + 4t − 3. Find the time when the acceleration is 14 ft/sec².

Answer 1

Displacement of a particle is given by

s = 2t³ − 5t² + 4t − 3  ...(I)

Differentiate w. r. t. t.

Velocity, v = `(ds)/dt`

= `d/dt (2t^3 - 5t^2 + 4t - 3)`

∴ v = 6t² − 10t + 4  ...(II)

Acceleration, a = `(dv)/(dt)`

= `d/dt(6t^2 - 10t + 4)`

∴ a = 12t − 10  ...(III)

Given, Acceleration = 14 ft/sec²

∴ 12t − 10 = 14

12t = 24

t = 2

So, the particle reaches an acceleration of 14 ft/sec² in 2 seconds.

Velocity, when t = 2 is

∴ v_{t = 2} = 6(2)² − 10(2) + 4

= 6(4) − 20 + 4

= 24 − 20 + 4

= 4 + 4

= 8 ft/sec

Displacement when t = 2 is

∴ s_{t = 2} = 2(2)³ − 5(2)² + 4(2) − 3

= 2(8) − 5(4) + 8 − 3

= 16 − 20 + 8 − 3

= −4 + 5

= 1 foot

Hence the velocity is 8 ft/sec and the displacement is 1 foot after 2 seconds.