Need help with a "motion and rates" question
- Thread starter Corro
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Hello, and welcome to FMH! 
The position \(x\) of an object undergoing constant acceleration is:
[MATH]x(t)=\frac{1}{2}at^2+v_0t+x_0[/MATH]
Let's orient our coordinate axis such that the car is initially at the origin, and the truck is at \(C\). And to the position of the car is:
[MATH]x_C(t)=\frac{1}{2}at^2+Vt[/MATH]
And the position of the truck is:
[MATH]x_T(t)=Vt+C[/MATH]
Now, the positions of the two vehicles will the same when they are at \(C+L\), and so we can find when that happens:
[MATH]\frac{1}{2}at^2+Vt=Vt+C[/MATH]
[MATH]t=\sqrt{\frac{2C}{a}}[/MATH]
And so we find, by using this time in either position functions (we'll use the truck for simplcity):
[MATH]V\left(\sqrt{\frac{2C}{a}}\right)+C=C+L[/MATH]
[MATH]a=2C\left(\frac{V}{L}\right)^2[/MATH]
Now, to find the speed of the car at this time/position, we may differentiate the car's position function:
[MATH]v_C(t)=x_C'(t)=at+V[/MATH]
Plug in for \(a\) and \(t\) to get:
[MATH]v_C\left(\sqrt{\frac{2C}{a}}\right)=\left(2C\left(\frac{V}{L}\right)^2\right)\left(\sqrt{\frac{2C}{a}}\right)+V=\left(2C\left(\frac{V}{L}\right)^2\right)\frac{L}{V}+V=\frac{2CV}{L}+V=V\left(1+\frac{2C}{L}\right)[/MATH]
Can you proceed?
The position \(x\) of an object undergoing constant acceleration is:
[MATH]x(t)=\frac{1}{2}at^2+v_0t+x_0[/MATH]
Let's orient our coordinate axis such that the car is initially at the origin, and the truck is at \(C\). And to the position of the car is:
[MATH]x_C(t)=\frac{1}{2}at^2+Vt[/MATH]
And the position of the truck is:
[MATH]x_T(t)=Vt+C[/MATH]
Now, the positions of the two vehicles will the same when they are at \(C+L\), and so we can find when that happens:
[MATH]\frac{1}{2}at^2+Vt=Vt+C[/MATH]
[MATH]t=\sqrt{\frac{2C}{a}}[/MATH]
And so we find, by using this time in either position functions (we'll use the truck for simplcity):
[MATH]V\left(\sqrt{\frac{2C}{a}}\right)+C=C+L[/MATH]
[MATH]a=2C\left(\frac{V}{L}\right)^2[/MATH]
Now, to find the speed of the car at this time/position, we may differentiate the car's position function:
[MATH]v_C(t)=x_C'(t)=at+V[/MATH]
Plug in for \(a\) and \(t\) to get:
[MATH]v_C\left(\sqrt{\frac{2C}{a}}\right)=\left(2C\left(\frac{V}{L}\right)^2\right)\left(\sqrt{\frac{2C}{a}}\right)+V=\left(2C\left(\frac{V}{L}\right)^2\right)\frac{L}{V}+V=\frac{2CV}{L}+V=V\left(1+\frac{2C}{L}\right)[/MATH]
Can you proceed?
