Is it possible to find when two cars will meet given this graph?

Question

The problem is as follows:

The figure from below shows the speed against time of two cars, one blue and the other orange. It is known that both depart from the same spot. Find the instant on seconds when one catches the other.

The given alternatives are:

$\begin{array}{ll} 1.&14\,s\\ 2.&16\,s\\ 3.&20\,s\\ 4.&25\,s\\ 5.&28\,s\\ \end{array}$

For this problem I attempted to do the "trick" using the areas behind the curves but I couldn't find the answer.

So far I could only state the equations as this: (I'm using $v_{r}=\textrm{orange car}$ and $v_{b}=\textrm{blue car}$

$v_{r}=8$

$v_{b}=t-5$

Since $v=\dfrac{dx}{dt}$

Then:

$\dfrac{dx}{dt}=8$

$x(t)=8t+c$

$x(0)=0\,, c=0$

$x(t)_{r}=8t$

$\dfrac{dx}{dt}=t-5$

$x(t)=\frac{t^2}{2}-5t+c$

$x(0)=0\, c=0$

$x(t)_{b}= \frac{t^2}{2}-5t$

So by equating both I could obtain the time isn't it?.

$8t=\frac{t^2}{2}-5t$

$0=\frac{t^2}{2}-13t$

$0=t(t-26)$

So time would be $26\,s$ But it doesn't seem to be in any of the alternatives given. Could it be that I'm not getting the right picture of this problem correctly?. Can somebody give me a help?.

Answer 1

The mistake in your computations is that for the second care you set $x(0)=0$. You should have $x(5)=0$

Note that for the second car you have $x'(t)=0$ when $0 \leq t \leq 5$ and $x'(t)=t-5$ when $t \geq 5$. You need to either start your computations at $t_0=5$ or treat this as a piecewise function.

Hint Alternatelly, the areas below the graphs represent the distance traveled. You want the two area below the graph to be equal.

It $t$ is the moment they meey, the area below the first graph is $A=8t$. The area below the second graph is $\frac{1}{2} (t-5)^2$. Make them equal.

Answer 2

Velocity is defined as the derivative of the distance function. Let $v_1(t)$ and $v_2(t)$ denote the velocity function for the first and second car respectively. Then, we can express the distance functions, $d_1(t)$ and $d_2(t)$ respectively, as an anti-derivative of the velocity function: $$d_1(t)=\int v_1(t)dt$$ and $$d_2(t)=\int v_2(t)dt.$$

To find when the two cars meet, find the values of $t$ that satisfy $d_1(t)=d_2(t)$, which can be done by integrating the velocity functions and then solving for $t$.

Answer 3

The orange car started moving $5$ seconds before the blue car started. By the time the blue car started moving, the orange car had already covered a $40$ meter distance. Therefore, we can say that:

Orange car distance $(d_o)=8x+40$

Blue car distance $(d_b)=\frac{x^2}{2}$

Now we make both the equations equal to each other, and solve for $x$ (time).

$8x+40=\frac{x^2}{2} \\ 16x+80=x^2 \\ 0=x^2-16x-80 \\ 0=(x-20)(x+4) \\ x=20,-4$

Since we can't have $-4$ seconds, we have to go with $20$ seconds. However, this is not the final answer. The time $20$ seconds just tells us how long it took after the blue car started moving for both distances to become equal. In reality, the orange car had already started moving $5$ seconds earlier, so we have to count that time as well. Therefore, the final answer is $25$ seconds.

If you wanted to take $t=0$ as the starting point for both cars (instead of $t=5$), then you would have to go with the advice of N.S. and write the blue car's velocity as a piecewise function.