How long does it take for the car to catch up to the truck on a straight track?

[Bry2323]

A truck and a car both start from rest at the same time on a straight track. The car is 25.0 m behind the truck at the start. The truck accelerates at 3.70 m/s^2 and the car accelerates at 4.40 m/s^2. How long does it take for the car to catch up to the truck?

Hey all, first post, didn't think it'd be so quick in this college physics class! Can someone help me figure this out...I have an idea that displacement of truck should equal the car plus 25.0 m right? Just having trouble getting there. Thank you!

 

[haruspex]

A truck and a car both start from rest at the same time on a straight track. The car is 25.0 m behind the truck at the start. The truck accelerates at 3.70 m/s^2 and the car accelerates at 4.40 m/s^2. How long does it take for the car to catch up to the truck?

Hey all, first post, didn't think it'd be so quick in this college physics class! Can someone help me figure this out...I have an idea that displacement of truck should equal the car plus 25.0 m right? Just having trouble getting there. Thank you!

Please don't delete the template. It's there to be filled in. What relevant equations do you know? Please show some attempt.

 

[Bry2323]

Sorry, didn't realize that until i searched around a little more.

So far I have displacement of truck equals 1/2a x t^2 = 1.85 m/s^2(t^2)
I am thinking displacement of car = displacement of truck + 25.0 m = 1/2at^2

I am left with 1.85 m/s^2(t^2) = 2.20 m/s^2(t^2) + 25.0 m?? This is where I am so very lost.

Thank you again.

 

[haruspex]

displacement of car = displacement of truck + 25.0 m

Which starts in front? What are the two displacements at t=0?

 

[Bry2323]

The truck starts in front. The displacement for the car is 0m and truck is 25m?

 

[haruspex]

The truck starts in front. The displacement for the car is 0m and truck is 25m?

Right, so what is your equation now?

 

[Bry2323]

I'm not sure, are you pointing out that displacement of truck should be displacement of car + 25.0 m?

 

[haruspex]

I'm not sure, are you pointing out that displacement of truck should be displacement of car + 25.0 m?

Yes. What you had before gives the wrong result at t=0.

 

[Bry2323]

I think I had it correct underneath that? I really don't know how to simplify that last line if that is the correct formula

 

[haruspex]

I think I had it correct underneath that? I really don't know how to simplify that last line if that is the correct formula

Your formula has the form At²+B=Ct². Can you not see how to simplify that?

 

[Bry2323]

I really can't, is that a quadratic? It's been awhile since I have been in Algebra.

 

[Bry2323]

ok I think i got it but I think I've done way too many steps

2.20 m/s^2(t^2) +25.0m = 1.85 m/s ^s (t^2)

.35 m/s^2(t^2) +25.0m = 0

.35(t^2) = -25.0s^2

t^2 = -71.43 s^2

t= 8.45 s

 

[haruspex]

I think I had it correct underneath that?

No, you didn't. You had

1.85 m/s^2(t^2) = 2.20 m/s^2(t^2) + 25.0 m

The left hand side has the truck's acceleration, so presumably represents the truck's displacement. So your equation says displacement of truck = displacement of car + 25m.
This is why you get an impossibility here:

t^2 = -71.43 s^2

t² cannot be negative. It is not valid simply to ignore the minus sign because it's inconvenient. In the present case, ignoring it does happen to give the right answer, because the minus sign should not have been there in the first place.

 

[Bry2323]

Ok well how do i move the 25.0 m to the right side without it being negative?

 

[haruspex]

Ok well how do i move the 25.0 m to the right side without it being negative?

This equation is wrong:

1.85 m/s^2(t^2) = 2.20 m/s^2(t^2) + 25.0 m

It does not correspond to your (correct) observation that "displacement of car = displacement of truck + 25.0 m". It has reversed it.
Correct the equation and the minus sign will go away.

 

[Bry2323]

ok perfect thank you, so what exactly am i saying when i say the displacement of car = displacement of truck +25.0 m, my brain is fried after today but I am having trouble understanding the theory behind it in lamens terms, my mind is wanting to think its the car +25 to make up for the deficit at the beginning but i know that's not right.

 

[haruspex]

what exactly am i saying when i say the displacement of car = displacement of truck +25.0 m

First, you need to make clear what you mean by that. It is correct if you mean their final displacements from their respective starting positions.
For each, its displacement from its starting position at time t is acceleration*t²/2.
So you have:
displacement_car = displacement_truck+25m
acceleration_car*t²/2 = acceleration_truck*t²/2+25m

 

[Bry2323]

I think I am confusing myselfs because I am thinking the only way the addition of 25m would need to be added was if the accelerations were the same, not sure why i can't grasp the idea, think I am just burnt out

 

[haruspex]

I think I am confusing myselfs because I am thinking the only way the addition of 25m would need to be added was if the accelerations were the same, not sure why i can't grasp the idea, think I am just burnt out

I think your confusion comes from not being clear in what you meant by displacement in "displacement of car = displacement of truck + 25.0 m".
If it means their respective displacements from their respective starting positions when they meet, it is true. If it means respective displacements from a common origin when they start, it is backwards.

 

[Bry2323]

Ok, I'll take a break and think it over, just trying to make sure I truly understand everything, you've been a big help, cheers!