Calculating Distance for Accelerating Objects

[Rumplestiltskin]

Homework Statement

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).

Homework Equations

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.

 

[Samy_A]

Homework Statement

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).

Homework Equations

3. The Attempt at a Solution [/B]
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.

How did you get 1.5t² = d?

 

[Rumplestiltskin]

How did you get 1.5t² = d?

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

 

[Samy_A]

It's the acceleration, multiplying by t² will give distance m.

Check your "SUVAT" equations.

 

[Rumplestiltskin]

Check your "SUVAT" equations.

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

 

[Samy_A]

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

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.

 

[Rumplestiltskin]

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.

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

1.5 * 16² = 384m.
12 * 16 = 192m.
?

 

[Samy_A]

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

1.5 * 16² = 384m.
12 * 16 = 192m.
?

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

 

[Rumplestiltskin]

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

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?

 

[Samy_A]

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?

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

 

[Rumplestiltskin]

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

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.

 

[Ray Vickson]

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

1.5 * 16² = 384m.
12 * 16 = 192m.
?

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

 

[Samy_A]

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.

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

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

 

[Rumplestiltskin]

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

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

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.

 

[Samy_A]

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.

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 traveled 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.

 

[Rumplestiltskin]

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 traveled 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.

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 traveled in that time. It seems arbitrary that it only outputs half the distance traveled in any time you input.

 

[Samy_A]

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 traveled in that time. It seems arbitrary that it only outputs half the distance traveled in any time you input.

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

Let's take the case of the motorcyclist, traveling 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.)