What is the velocity and position of this problem?

Question

A particle moves along the x axis with its position relative to the origin and has an acceleration function a(t) = 4(t) + 6, in cm/s^2 and time t in seconds. If the initial velocity of the particle is -5cm/s and its initial position is 8cm right of the origin, what is the velocity and position when t=2 seconds?

Answer 1

Hi there, I hope this helps:

1. If a(t)=4t+6 is your acceleration vector, then the antiderivative of this function would be you velocity function.
2. The antiderivative of a(t)=4t+6 is v(t)=2t²+6t+c by the power rule of integration. To solve for c, we can use the information given in the problem (at time 0, the velocity is -5).

-5=2(0)²+6(0)+c

-5=c

Now, plugging in c, our velocity equation is v(t)=2t²+6t-5

To find the velocity at t=2, we plug in 2 to the above equation

v(2)=2(2)²+6(2)-5

v(2)=8+12-5

v(2)=15 cm/s

3.Next, to find the position of the particle, we must derive the position function.

The position function is the antiderivative of the velocity function v(t)=2t²+6t-5

By the power rule of integration,

p(t)=(2/3)t³+3t²-5t+c

To solve for c, we use the information given in the problem (at t=0, the particle's position is at 8cm).

8=(2/3)(0)³+3(0)²-5(0)+c

8=c

Plugging in c:

p(t)=(2/3)t³+3t²-5t+8

So the position of the particle at t=2 is p(2):

p(2)=(2/3)(2)³+3(2)²-5(2)+8

p(2)=(16/3)+12-10+8

p(2)=(16/3)+10

p(2)=46/3 cm