Kinematics - finding the time when two moving objects meet

[yurono]

Homework Statement

Red kangaroos are known to reach speeds of nearly 11.0 m/s. If a red kangaroo is 52 m away from the bus, and the bus starts to accelerate away from the kangaroo at a rate of 1.1 m/s², does the kangaroo catch the bus?

Relevant Equations

d = v*t + 1/2(a*∆t²)

I tried doing (vi)(∆ t) / 1/2(a)(∆ t) = 1/2(a)( ∆ t)² / 1/2(a)(∆ t)
That allows you to cross out the two accelerations out and you end with an answer of 20. But apparently, they never meet. I then tried making them into two different equations and using graphing technology (Desmos) and they don't meet, as one is a parabola (acceleration) and one is linear (velocity), but I will not have access to that when it comes to tests. How would I go on solving this?

 

[gneill]

Homework Statement:: Red kangaroos are known to reach speeds of nearly 11.0 m/s. If a red kangaroo is 52 m away from the bus, and the bus starts to accelerate away from the kangaroo at a rate of 1.1 m/s², does the kangaroo catch the bus?
Relevant Equations:: d = v*t + 1/2(a*∆t²)

I tried doing (vi)(∆ t) / 1/2(a)(∆ t) = 1/2(a)( ∆ t)² / 1/2(a)(∆ t)

Can you elaborate on our reasoning for that last equation?

 

[yurono]

Can you elaborate on our reasoning for that last equation?

Our teacher wrote this to solve a similar question. If you cancel the 1/2 (a)(∆t), you'd be left with ∆t since the other side has ∆t²

 

[gneill]

Our teacher wrote this to solve a similar question. If you cancel the 1/2 (a)(∆t), you'd be left with ∆t since the other side has ∆t²

Sorry, I'm unable to motivate the term 1/2 (a)(∆t). Where does that come from?

How "similar" was the other question to this one?

Can you provide a derivation for the equation? Simply copying an equation from a different problem is often a really bad idea unless you have a really good reason why it should apply to the current problem. Do you have such reason?

I think your best approach would be to consider this problem separately and to employ basic SUVAT equations. This approach will not fail you when problem details differ from what you have previously encountered.

 

[yurono]

The other question was this: 2 lizards are 50 meters apart. Lizard A starts walking towards Lizard B at a constant speed of 1.57 m/s. At the same instant, Lizard B accelerates uniformly from rest at a rate of 0.25 m/s² . How long before the two lizards meet?

As both questions had a d value, an acceleration value, and velocity value, I thought it'd make sense to use the same process to this question. When you say SUVAT, are you referring to the five kinematic equations? Wouldn't d = v*t + 1/2(a*∆t²) work since it has all (or well, most) of the givens?

 

[gneill]

The other question was this: 2 lizards are 50 meters apart. Lizard A starts walking towards Lizard B at a constant speed of 1.57 m/s. At the same instant, Lizard B accelerates uniformly from rest at a rate of 0.25 m/s² . How long before the two lizards meet?

Yes, it is a similar question, but I can't grasp how the terms of the equation that was presented were arrived at.

As both questions had a d value, an acceleration value, and velocity value, I thought it'd make sense to use the same process to this question. When you say SUVAT, are you referring to the five kinematic equations? Wouldn't d = v*t + 1/2(a*∆t²) work since it has all (or well, most) of the givens?

Yes, that is what I was referring to with regards to SUVAT. And yes, that equation is applicable. Express the positions of the kangaroo and bus (with a suitably chosen origin) in terms of basic SUVAT equations. Then see if they meet.

 

[kuruman]

You don't need graphing technology. A good strategy and algebra to implement it is all you need.

This is known as a "catch-up" problem that is solved by requiring that the two objects be at the same position at the same time. The general strategy for solving such problems is
1. Write a SUVAT equation giving the position of object 1 at any time t.
2. Write a second SUVAT equation giving the position of object 2 at any time t.
3. Say with an equation that at specific time t_c (the catch-up time) the two objects are at the same position.
4. Solve that equation for t_c.
5. Use t_c to find anything else the problem might ask.

In this case if a solution for t_c exists, you will have found the catch-up time. If a solution does not exist, the kangaroo will never catch up.

 

[Orodruin]

Sorry, I'm unable to motivate the term 1/2 (a)(∆t). Where does that come from?

It is just an arbitrary division to solve for $\Delta t$ from the (wrong) expression equating $v \Delta t$ to $a\Delta t^2/2$.

1. Write a SUVAT equation giving the position of object 1 at any time t.
2. Write a second SUVAT equation giving the position of object 2 at any time t.
3. Say with an equation that at specific time tc (the catch-up time) the two objects are at the same position.

This is, in essence, what the OP has already done, but with an incorrect expression for the positions as the initial distance has not been considered.

 

[kuruman]

This is, in essence, what the OP has already done, but with an incorrect expression for the positions as the initial distance has not been considered.

True, but OP's posts so far are not convincing that OP understands the justification behind the method. OP wrote down a single equation and then got an equality of two expressions by borrowing elements from a solution to another problem apparently without paying attention to any differences between the two situations. My thinking was that by outlining the general strategy OP would be able to see what the two situations have in common and what is different between them, the initial distance being (perhaps) one of the differences.

 

[archaic]

Here is what you have:
The bus, which is $52\mathrm m$ away from the kangaroo is trying to increase that distance as it is accelerating. The kangaroo, on the other hand, is trying to decrease that distance which separates it from the bus and, ultimately, reach the vehicle (and you are asked to figure out if he can).
Initially you have $x_b-x_k=52$, but as they start moving $x_b-x_k=52+(\text{distance the bus has moved})-(\text{distance the kangaroo has moved})$.
You need to, first, find how you can establish the expression for the distance traveled by both the animal and the vehicle, and then think what it means for the kangaroo to have reached the bus.
Here is the equation you have been given more clearly written: $d=vt+\frac{1}{2}at^2$.