Question

. A car moves in a straight line such that for a short time its velocity is defined by
v = (5t2 + 3t) ft/s, where ‘t’ is in seconds. Determine its position and acceleration when t = 4s. When t = 0, s = 0.

Answer 1

It is given that a car moves in a straight line such that its velocity is defined by v = (5t² + 3t) ft/s where t is in seconds.

we have to find the position and acceleration of car when t = 4s where at t = 0, s = 0

we know, rate of change of displacement with respect to time is velocity.

v = (5t² + 3t)

⇒ ds/dt = (5t² + 3t)

⇒∫ds = ∫(5t² + 3t)dt

⇒s = [5t³/3 + 3t²/2]

⇒s = 5(4)³/3 + 3(4)²/2 = 320/3 + 24

= 106.67 + 24 = 130.67 ft

therefore, displacement is 130.67 ft

we know, rate of change of velocity with respect to time is known as acceleration.

so, a = dV/dt = d(5t² + 3t)/dt = 10t + 3

now putting t = 4 sec

so, a = 10 × 4 + 3 = 43 ft/s²

therefore, acceleration is 43 ft/s²