A cricket jumps in the air with a velocity of $3.3\;m/s$. Its height (in meters) after $t$ seconds is given by $h(t) = 3.3 t - 4.9 t^2$.
How fast is the cricket moving after $t = 0.1$ seconds?
A cricket jumps in the air with a velocity of $3.3\;m/s$. Its height (in meters) after $t$ seconds is given by $h(t) = 3.3 t - 4.9 t^2$.
How fast is the cricket moving after $t = 0.1$ seconds?
Find $h'(t) = \dfrac {dh}{dt}$, which measures the instantaneous velocity of the cricket at time $t$.
Then evaluate this derivative at time $t = 0.1$.
Hint The derivative of position (in this case the cricket moves vertically, so position means height) is velocity...