I'm actually getting stuck with a tricky part of a math problem using a quadratic function.
An object propelled with an initial speed, has its altitude given after $t$ seconds by the quadratic function $h$ (in meters).
$$h(t) = -5t² + 10t + 15$$
The time when the object touches the ground is called $t_0$.
$$t_0 = 3$$
The problem is that I have to calculate the distance travelled by the object between $t=0$ and $t_0$. I don't know what to do while I haven't got the value of the object's speed.
After this, I'm asked to calculate the object's speed during its rise and during its fall.
I specify that I have no graphic of the function.
What shall I do?
Thanks for your answers
Best Answer
Remember that the velocity is the time derivative of the displacement, so we can say that the velocity as function of time is :
$$v(t)=-10t+10$$ Note that the velocity is negative for $t>1$ and is positive for $t<1$. This implies that the object changed its direction during the time interval $(t=0$ to $t=3)$. What it means for us is that we cannot just calculate the change in $h$ and say that it is the distance travelled.
We will need to divide the motion into 2 intervals- one is $(t=0$ to $t=1)$ and the other is $(t=1$ to $t=3)$, and add up the absolute values of their individual displacements to get the distance travelled. Doing so, you can easily obtain the answer as 25 m.
As for the object's speed, it is not constant, but you can give a time function for it, which will simply be the absolute value of velocity.
$$ s(t) = \left\{ \begin{array}{lr} 10(1-t) & : t\le1\\ 10(t-1) & : t\ge1 \end{array} \right. $$
This answer assumes that the motion is in only 1-D