How do you find values of t in which the speed of the particle is increasing if he position of a particle moving along a line is given by
#s(t)=2t^3-24t^2+90t+7# for #t≥0#?

Question

Charles Arocha · Accepted Answer

If #s(t) = 2t^3 − 24t^2 + 90t + 7#
The speed (at time t) is given by
#s'(t) = 6t^2 -48t +90#

Acceleration is given by the second derivative
# s''(t) = 12 t - 48#

Critical point for acceleration (when the speed changes from decreasing to increasing) occurs when
#s''(t) = 12t - 48 = 0#

That is when #t = 4#

This might be easier to see if we consider a table of time(t) and speed at time t:

We can see that the speed actually decreases until somewhere between #t=3# and #t=5# when it starts to increase again.

Ezekiel Alexander · Answer

undefined To find the values of $ t $ in which the speed of the particle is increasing, you need to determine when the acceleration $ a(t) $ is positive. The acceleration is the derivative of velocity, which in turn is the derivative of position.

1. First, find the velocity function $ v(t) $ by taking the derivative of the position function $ s(t) $.
   $ v(t) = \frac{ds}{dt} = 6t^2 - 48t + 90 $

2. Next, find the acceleration function $ a(t) $ by taking the derivative of the velocity function $ v(t) $.
   $ a(t) = \frac{dv}{dt} = 12t - 48 $

3. Set $ a(t) $ greater than zero and solve for $ t $.
   $ 12t - 48 > 0 $
   $ 12t > 48 $
   $ t > 4 $

Therefore, the speed of the particle is increasing for $ t > 4 $.

How do you find values of t in which the speed of the particle is increasing if he position of a particle moving along a line is given by #s(t)=2t^3-24t^2+90t+7# for #t≥0#?