The position function of a moving particle is given by p(t)=t2−9t+13 feet at t seconds. Find all points in time where the particle’s speed is 5ft/s.

Question

Separate multiple answers with a comma. If the particle never reaches this speed, write None for your answer.

Answer 1

Answer:

P(t) = t² - 9t + 13 . First find the velocity vector that can be found as V(t) = dP(t) / dt or the derivative of the position vector wrt to time

V(t) = 2(t)^2-1 - 9(t)^1-1 + 0 = 2t¹ -9t⁰ = 2t -9(1) = 2t -9 Substituting the particle speed

V(t) = 5ft/s = 2t - 9 , 2t = 5 +9 = 14 and t = 14 / 2 = 7s so there is only one time at 7 sec when the particle speed is 5 ft/s, because the equation of velocity vector is a straight line and there is only one value of speed for each point in time. Also note that the position vector is a curve or parabola and the derivative of it is the slope of the tangent point to the curve and the velocity at time t = 7 sec

if the velocity was zero then 0 = 2t -9 , and 2t = 9, t = 9/2 = 4.5 s different from 7 s this illustrates that there is only one value for speed for any value of time.

Answer 2

Hi Kiley,

If the position function is given as p(t) = t^2 - 9t + 13 how do we find the speed?

We take the derivative of the position function.

So, p'(t) = 2t - 9

This function tells us the speed of a moving particle at a given time.

In this case, we want to know at what time does the particle reach 5 m/s?

We want p'(t) = 5 = 2t - 9

Now, solve for t: 5 = 2t -9

Can you solve for t?