Question

Proposed Kinematic Exercise I

Consider a body moving on a straight path with a speed given by v=(3t^2-6t) m/s, where t is in seconds. If the body is at x=4m when t=0s, determine (a) its position at t=4s and (b) its acceleration at t=2s. (c) What is the direction of body movement between t=0s and t=4s? Justify your answer.

Answer 1

(a) 20 m

(b) 6 m/s²

(c) Between t=0 and t=2, the body moves to the left.

Between t=2 and t=4, the body moves to the right.

Step-by-step explanation:

v = 3t² − 6t

x(0) = 4

(a) Position is the integral of velocity.

x = ∫ v dt

x = ∫ (3t² − 6t) dt

x = t³ − 3t² + C

Use initial condition to find value of C.

4 = 0³ − 3(0)² + C

4 = C

x = t³ − 3t² + 4

Find position at t = 4.

x = 4³ − 3(4)² + 4

x = 20

(b) Acceleration is the derivative of velocity.

a = dv/dt

a = 6t − 6

Find acceleration at t = 2.

a = 6(2) − 6

a = 6

(c) v = 3t² − 6t

v = 3t (t − 2)

The velocity is 0 at t = 0 and t = 2. Evaluate the intervals.

When 0 < t < 2, v < 0.

When t > 2, v > 0.