An object moving in a straight line with an initial velocity of 10 m/s undergoes an acceleration of a ( t ) = 6 t − 10 m/s 2 , t seconds after an experiment begins.

Question

An object moving in a straight line with an initial velocity of  m/s undergoes an acceleration of  ,  seconds after an experiment begins.

The velocity of the object after seconds is 

The position of the object after seconds is 

Answer 1

dv/dt = a so dv = a dt

∫dv = ∫a dt

∫dv = ∫(6t - 10) dt

V = 3t² - 10t + C₁

We are told that the initial velocity = 10 meaning when t = 0, V = 10 so:

10 = 3(0)² - 10(0) + C₁

C₁ = 10

Therefore the velocity function is v(t) = 3t² - 10t + 10

ds/dt = v so ds = v dt

∫ds = ∫v dt

∫ds = ∫(3t² - 10t + 10) dt

s = t³ - 5t² + 10t + C₂

We are told to write a function of position relative to the starting point so we can say that at t = 0, s = 0.

So 0 = t³ - 5t² + 10t + C₂ or C₂ = 0

So s(t) = t³ - 5t² + 10t

Answer 2

v(t) = ∫a(t)dt = 3t²-10t + v(0) = 3t²-10t+10 m/s

s(t) = ∫v(t)dt = t³-5t² + 10t + s(0) = t³-5t² + 10t + 0 = t³-5t² + 10t m from starting point