Calculating speed and distance travelled by an object using quadratic function

Question

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

Answer 1

Based upon the information that you have given, I would assume that the problem is a 1D problem (i.e. a gun shot straight up in the air). This actually makes it easier. You are correct, if you had the projectile's speed, you could compute the arc length using the formula: $$\int_0^3 \sqrt{1+v(t)^2}dt$$ Note that we can find the projectile's speed. We need to differentiate the position formula, which will give us velocity. Since the arc length formula squares the velocity term, the signs don't matter. Therefore, let us differentiate the position formula to obtain the velocity: $$\frac{d}{dt} \left(-5t^2+10t+15\right)=-10t+10=v(t)=10(1-t)$$ Now that we have the velocity, we can find the speed of the object at any point in time. The difference between speed and velocity is that speed is a scalar, while velocity is a vector. In the 1D case, this changes to speed being the absolute value of velocity.

Let us now compute the arc length: $$\int_0^3 \sqrt{1+(10(1-t))^2}dt$$ $$\int_0^3 \sqrt{1+100(1-2t+t^2)}dt$$ $$\int_0^3 \sqrt{100t^2-200t+101}dt$$ Notice how hard this integral is getting. For reference, Wolfram Alpha gives the numerical output of that integral to be: $$\approx 25.38$$ There is an easier way to do it. We know that the projectile has positive velocity (up in the air) at first, then it changes to negative velocity (Towards the ground) until it impacts the ground. We can find the total distance travelled by finding how far it goes up in the air, plus how far it falls to the ground: $$distance=|h(0)-h(t_{peak})|+|h(t_{peak})-h(3)|$$ The $t_{peak}$ value is where the velocity reaches 0. Trivial solution: $t=1$. Therefore, the equation becomes: $$distance=|h(0)-h(1)|+|h(1)-h(3)|$$ $$distance=|15-20|+|20-0|$$ $$distance=25$$

Answer 2

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