# Distance / speed / time

1. Jan 20, 2016

### Rumplestiltskin

1. The problem statement, all variables and given/known data
A car passes through traffic lights as they turn from red to green at a speed of 12ms-1. A motorcyclist, starting from rest, accelerates from the traffic lights in the same direction as the car at 1.5ms-2. After what distance from the lights will they meet? (OPTIONS: 16m, 120m, 168m, 192m).

2. Relevant equations
3. The attempt at a solution
12t = d, and 1.5t2 = d.
12t = 1.5t2.
1.5t2 - 12t = 0.
t(1.5t -12) = 0.
t = 0 or 8.

12ms-1 * 8s = 96m. Not an option.

2. Jan 20, 2016

### Samy_A

How did you get 1.5t2 = d?

3. Jan 20, 2016

### Rumplestiltskin

It's the acceleration, multiplying by t2 will give distance m.
1.5 * 82 = 96m
12 * 8 = 96m

4. Jan 20, 2016

### Samy_A

Check your "SUVAT" equations.

5. Jan 20, 2016

### Rumplestiltskin

s = 0.5 * 1.5 * 82 = 48m? Why wouldn't that work in its own right?
48 / 12 = 4s, which is inconsistent.

6. Jan 20, 2016

### Samy_A

Forget that 8.

For the motorcyclist, you now seem to use the equation d=0.5*a*t². That is the correct equation. Now go back to your first post, and use this instead of d=a*t² as you did there.

7. Jan 20, 2016

### Rumplestiltskin

12t = 0.75t2
0.75t2 - 12t = 0
t(0.75t - 12) = 0
t = 0 or 16.

1.5 * 162 = 384m.
12 * 16 = 192m.
???

8. Jan 20, 2016

### Samy_A

Almost there. I don't understand what 1.5 * 162 means in this context.
12*16 is the distance travelled by the car as they meet again. Obviously the distance should be the same for the motorcyclist. What was that equation again?

9. Jan 20, 2016

### Rumplestiltskin

It's t2 multiplied by the acceleration of the motorcycle. Since the acceleration is m/s2, shouldn't that function give you a value of some significance?

10. Jan 20, 2016

### Samy_A

I don't know if it has any significance, but it is not relevant to the problem.
The distance travelled by the motorcyclist is, as you said yourself, and applied in 12t = 0.75t², given by 0.5*a*t².

11. Jan 20, 2016

### Rumplestiltskin

This seems to go against the fundamentals.
If the motorcycle travels a certain distance after time t, t should give you that distance when plugged into ms-2. The seconds squared cancels out. Otherwise the units don't make sense.

12. Jan 20, 2016

### Ray Vickson

You forgot that pesky factor "1/2" again!

13. Jan 20, 2016

### Samy_A

The correct equation is d=0.5*a*t². The 0.5 has no units.

Recapitulating:
the car has travelled 12*16 m = 192 m when the two meet again.
The motorcyclist obviously has travelled the same distance when they meet again. You could doublecheck this by using the correct formula for the motorcyclist: d=0.5*a*t².

Last edited: Jan 20, 2016
14. Jan 20, 2016

### Rumplestiltskin

I'm baffled. They will meet at the same time, which is 16s. Therefore plugging this into the acceleration should give me the same distance of 192m. Otherwise the units do not work.

15. Jan 21, 2016

### Samy_A

You are baffled because you don't write down the equations.
Using your terminology, we have:
for the car, driving at constant speed $v=12\ m/s$: $d_c(t)=vt$.
For the motorcyclist, driving from rest at constant acceleration $a=1.5\ m/s²$: $d_m(t)=\frac{1}{2}at²$.

As you are looking for the distance at which they meet, the equation to solve is $d_c(t)=d_m(t)$.
You solved that part correctly, getting 16 seconds for $t$.
Now all you have to do is plug in that value, 16, in the equation you have for $d_c$ or for $d_m$. You are right, they both should give the same result, and they do.

Looking at dimensions is a good method to check whether a result makes sense. If you are computing a distance, and find a value in seconds, you know something went wrong.
But dimensions being correct is a necessary condition, not a sufficient one.
If a is an acceleration, and t a time, at² will be a value in meters. That is correct. But so will 10at², or 100at². All the products of at² with a dimensionless constant will give a value in meters. There is only one that gives you the distance travelled from rest when a is a constant acceleration: $\frac{1}{2}at²$. The "pesky" ½ Ray mentioned.

That's why the exercise template contains a section "Relevant equations". Had you looked them up, written them down, and used them in the calculation, you would have solved this exercise in post 1.

Last edited: Jan 21, 2016
16. Jan 21, 2016

### Rumplestiltskin

I'm familiar enough with suvat, I just wasn't expecting to use it in this question, partly because of its mark count and partly because this is the first instance I've needed to use it outside of projectile motion.

Thanks for the explanation. It's beginning to make sense, though I'm still shaken up. I thought it was a given that plugging values into any unit will always give you the corresponding values; if you input the time in seconds for ms-2, you will arrive at the metres travelled in that time. It seems arbitrary that it only outputs half the distance travelled in any time you input.

17. Jan 21, 2016

### Samy_A

It's not arbitrary, it follows from the definitions of velocity and acceleration.

Let's take the case of the motorcyclist, travelling from rest in a straight line with constant acceleration a.
By definition, the acceleration is the rate of change of the velocity with respect of time.
So $\frac{dv}{dt}=a$, which gives $v(t)=v_0 +at$. For the motorcyclist $v_0=0$, so $v(t)=at$.
Similarly, velocity is the rate of change of position with respect of time.
This give $\frac{ds}{dt}=v(t)=at$. Integrating this then gives $s(t)=s_0+\frac{1}{2}at²$. There you have that ½, a direct consequence of two basic definitions.
For the motorcyclist, $s_0=0$, so we finally get $s(t)=\frac{1}{2}at²$.

(The more general 3-d equations are derived similarly.)