How Long Does It Take for a Car to Pass a Truck After Reaching the Speed Limit?

[kaylamalanie]

Homework Statement

A car moving at a velocity of 20 m/s is behind a truck moving at a constant velocity of 18 m/s. When the car is 50 m behind the front of the truck, the car accelerates uniformly at 1.8 m/s^2. The car continues at the same acceleration until it reaches a velocity of 25 m/s, which is the legal speed limit. The car then continues at a constant velocity of 25 m/s, until it passes the front of the truck. Find the time interval from the point that the car reaches the speed limit until it passes the truck.

Homework Equations

v=v0+at^2 is what I used, but now I'm unsure.

The Attempt at a Solution

I'm doing test corrections on a test. I know that the answer is approx. 5.2 s, but I'm really stuck at getting to it. Using the above equation, I got 1.97 s, but the answer isn't right or reasonable. I'm really stuck.

 

[berkeman]

Homework Statement

A car moving at a velocity of 20 m/s is behind a truck moving at a constant velocity of 18 m/s. When the car is 50 m behind the front of the truck, the car accelerates uniformly at 1.8 m/s^2. The car continues at the same acceleration until it reaches a velocity of 25 m/s, which is the legal speed limit. The car then continues at a constant velocity of 25 m/s, until it passes the front of the truck. Find the time interval from the point that the car reaches the speed limit until it passes the truck.

Homework Equations

v=v0+at^2 is what I used, but now I'm unsure.

The Attempt at a Solution

I'm doing test corrections on a test. I know that the answer is approx. 5.2 s, but I'm really stuck at getting to it. Using the above equation, I got 1.97 s, but the answer isn't right or reasonable. I'm really stuck.

Welcome to the PF.

Please show us your detailed calculations, so that we can give you a hand.

 

[Sourabh N]

I wonder how you got 1.97 s. What values did you put in?

Also can you spell out your approach? Sometimes it helps to write out what you're thinking :)

 

[kaylamalanie]

25 m/s=18m/s+(1.8 m/s^2)t^2
7m/s=(1.8m/s^2)t^2
3.89s=t^2
1.97=t
that's what I did. I'm not sure if I used the wrong numbers or what.

 

[kaylamalanie]

I think the two velocities may have confused me. I put 18 m/s in as the initial velocity and 25 m/s in as the constant velocity. 1.8 m/s^2 was my acceleration.

 

[Sourabh N]

That is the time the car takes to accelerate from 18 m/s to 25 m/s. The question wants you to calculate how much time, from this instant, does the car take to overtake the truck.

 

[kaylamalanie]

oh, that explains why my teacher wrote I was half through. do I then take that number and plug it into the same equation?

 

[kaylamalanie]

hang on, the initial velocity is 20 m/s i believe, it's the truck with the 18 m/s. so if i correct that, do I plug it in?

 

[Sourabh N]

My bad. There is an error in your previous calculation. The car is moving at 20 m/s, not 18.

Now the situation is this. The car is moving at 25 m/s, the truck at 18 m/s, and they are some distance apart. How much time would it take for the car to overtake?

 

[kaylamalanie]

well it would take the car 1.67 s to accelerate from 20m/s to 25 m/s. this is where i get stuck. I tend to have trouble understanding what to do with the pieces of info I find. I understand the concept of trying to find the equation with the variable you're looking for to solve for, but I don't understand the relationship between the cars.

 

[Sourabh N]

Ok look at it like this. You know how far apart they are, initially. You should be able to calculate how far apart they will be when the car has reached 25 m/s speed. After that, you need to find how much more time it takes to cover the gap.

Which of these statements do you have trouble with?

 

[kaylamalanie]

calculating how far apart they are when the car has reached 25 m/s.

 

[Sourabh N]

Can you find distance traveled by the car while accelerating? And distance traveled by the truck in the meantime?

 

[kaylamalanie]

I got 52.5 m for the car. I don't understand how to find it for the truck.

 

[Sourabh N]

Well the truck is moving at a constant velocity. So the distance it travels is velocity*time. Time is the time car took to accelerate. You can find that?

 

[Sourabh N]

Also, maybe its a typo or you did a mistake in calculation, but the car moves by 62.5 m.

 

[kaylamalanie]

30.1 m for the truck. I knew distance equals velocity*time, but I have trouble relating things.

 

[Sourabh N]

Happens :) So, now you know how far apart they are when the car has reached 25 m/s. Read my earlier post again, about what comes next.

 

[kaylamalanie]

It wasn't a typo, I used d=v0+.5at^2, is the equation wrong?

 

[kaylamalanie]

they would be 32.4 m apart?

 

[Sourabh N]

It wasn't a typo, I used d=v0+.5at^2, is the equation wrong?

That is right. And looks like you did a mistake in calculating the time the car takes to accelerate.

VLaTeX Code: _{f} = VLaTeX Code: _{i} + a*t

VLaTeX Code: _{f} = 25 m/s
VLaTeX Code: _{i} = 20 m/s
a = 1.8 m/sLaTeX Code: ^{2}

So t is?

 

[kaylamalanie]

2.78 s. I made the mistake because on my test my teacher said the t in the formula was squared which I didn't think it was, just confusing me more.

 

[kaylamalanie]

im referring to the t in the vf=vi+at. it was what i initially used.

 

[Sourabh N]

Good! Now, find the correct distance the car and the truck move.

 

[kaylamalanie]

using d=v0+.5at^2?

 

[Sourabh N]

Yes. Only the formula is really d = v0*t + 0.5*a*t^2

 

[kaylamalanie]

I finally got 62.5 m for the car! would it be 50.04 m for the truck?

 

[Sourabh N]

Yes, great! Now, you know how far apart they were initially and how much each of them moved. You can calculate how far apart they are now?

 

[kaylamalanie]

would they be 12.5 m apart when the car reached 25m/s?

 

[Sourabh N]

No. That is the *change* in their relative distance.

Initially they were 50 m apart. The car moves by 62.5 m, the truck by 50 m. How far apart they are now?
It would help if you draw a figure, how the system looks initially and how it looks finally.

 

[kaylamalanie]

37.5 m? if the car starts at the origin, then its final position would be 62.5m and the truck starts at 50, its final position would be 100 m

 

[Sourabh N]

Perfect. Now? :)

 

[kaylamalanie]

im not sure which equation to use.

 

[Sourabh N]

Ok you know the distance between the car and truck, and you know their speeds and you need to find the time.

In terms of relative velocity, its just distance = relative velocity*time

But in case you're not familiar with relative velocity, look at it like this. Assume the car takes time T to overtake the truck. It moves a distance 25*T during this. The truck moves a distance 18*T in the meantime. Now, The car should move 37.5 m more than the truck to overtake it.