How do you find the instantaneous acceleration for the particle whose position at time #t# is given by #s(t)=3t^2+5t# ?

Question

Emily Ammerman · Accepted Answer

I would derive twice your expression for the particle position (with respect to time).
#(ds)/dt=6t+5#
This is the instantaneous velocity #v#;
Deriving again you get:
#(dv)/dt=6# which is your instantaneous (and constant) acceleration. Remember that the derivative with respect to time tells you how your quantity changes with time.
The first derivative of position then tells you how position changes with time (velocity). Deriving velocity tells you how velocity changes in time, which is acceleration.

Paisley Johnson · Answer

undefined To find the instantaneous acceleration for the particle whose position at time $ t $ is given by $ s(t) = 3t^2 + 5t $, we differentiate the position function $ s(t) $ twice with respect to time $ t $ to obtain the acceleration function.

First, we find the velocity function $ v(t) $ by differentiating $ s(t) $ once with respect to $ t $:

$$ v(t) = \frac{ds}{dt} = \frac{d}{dt}(3t^2 + 5t) $$

$$ v(t) = 6t + 5 $$

Then, we find the acceleration function $ a(t) $ by differentiating $ v(t) $ with respect to $ t $:

$$ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t + 5) $$

$$ a(t) = 6 $$

Therefore, the instantaneous acceleration for the particle is constant and equal to $ 6 $ units per second squared.