Question

In deep space (no gravity), the bolt (arrow) of a crossbow accelerates at 141 m/s^2 and attains a speed of 108 m/s when it leaves the bow.

1. For how long is it accelerated?
Answer in units of s

2. What speed will the bolt have attained 4.8 s after leaving the crossbow?
Answer in units of m/s

3. How far will the bolt have traveled during the 4.8 s? Answer in units of m

Answer 1

Hi!

\[\LARGE 1.\]Here's the situation; you have the arrow, and it's at rest (with you and the bow). Then the bolt gets accelerated! Acceleration is just a change in velocity, and that's what's happening.
As the bolt accelerates at \(141\ m/s^2\) like the problem says, it gets faster. After a time, it is released, and has reached a velocity of \(108\ m/s\).
Average acceleration is the change in velocity divided by the change in time while the acceleration was applied.\[a=\frac{\Delta v}{\Delta t}=\frac{v_{final}-v_{initial}}{\Delta t}\]The \(\Delta\) indicates the change of what that variable is used for. You want the change in time. So you do algebra, multiplying both sides by \(\Delta t\) and dividing both sides by \(a\). Then you plug in your numbers, units will match up, and you'll have your answer 1.

Answer 2

When you find that answer, it will help with the rest!

Answer 3

Hi! Thank you so much! But I'm still a little confused! What value should I plug for a.

Answer 4

\(a\) is the variable for acceleration, and that means the acceleration of the bolt, for this problem. The question tells you the bolt's acceleration.

Answer 5

\(a=141\ m/s^2\)

Answer 6

\[\Huge \color{blue}{\ddot\smile}\]

Answer 7

I'm so sorry, but I'm still a little confused! I know a = 141 m/s^2, however, I don't know how to follow up the problem.

Answer 8

For problem 1, you mean?

Answer 9

Yes.

Answer 10

I see! When you read my first response, where do I lose you?

Answer 11

As you told me, I first multiply both sides by Δt and then divided both sides by a, but I don't know if I did right. Also, I'm kind of confused with the values I have to plug in (except a = 141 m/s^2).

Answer 12

I first multiplied*

Answer 13

That's alright! Then the algebra needs to be handled first! The starting expression is commonly seen, and is just the way to define acceleration with math instead of English.

\[a=\frac{v_{final}-v_{initial}}{\Delta t}\]
After multiplying:\[a\ \Delta t=\frac{\left(v_{final}-v_{initial}\right)\ \Delta t}{\Delta t}\]
\(\Delta t\) divided by \(\Delta t\) is 1, like any number divided by itself. So they cancel out (multiplying by 1 doesn't change anything, so you can take those \(\Delta t\)s away).
\[a\ \Delta t=\frac{\left(v_{final}-v_{initial}\right)\ \cancel {\Delta t}}{\cancel{\Delta t}}=a\ \Delta t = v_{final}-v_{initial}\]Now we see that\[a\ \Delta t = v_{final}-v_{initial}\]

Answer 14

I'm starting to see!

Answer 15

Then you divide by \(a\). This is more algebra, but it's worth it when you find the answer :)

\(a\ \Delta t = v_{final}-v_{initial}\qquad\) from before
Divide both sides by \(a\) and you'll get\[\frac{a\ \Delta t }{a}=\frac{ v_{final}-v_{initial}}{a}\]\(\large \frac{a}{a}\normalsize =1\), and multiplying by 1 changes nothing. So you can take those \(a\)'s out.\[\frac{\cancel a\ \Delta t }{\cancel a}=\Delta t=\frac{ v_{final}-v_{initial}}{a}\]

There, you have solved for \(\Delta t\).\[\Delta t=\frac{ v_{final}-v_{initial}}{a}\]
As I said when we rand through the scenario, the bolt starts at rest, its velocity is 0. That is its "initial velocity," or \(v_{initial}\).

For the final velocity, you look for its speed as it stops being accelerated by the crossbow. The problem gives you that, as well.

Answer 16

So, "initial velocity" = 0, "final velocity" = 108 m/s, and "a" = 141 m/s^2, right?

Answer 17

Yup!

Answer 18

So, should my final answer be 0.765957 s?

Answer 19

Yep!

Answer 20

I have to go, sorry. I'll post more later, even if you're not online.
Let me know what you get or what you think as you continue working!
I'll tell you that #2 is simpler than it might seem at first!

Answer 21

Good luck! :)

Answer 22

You are such a great helper!—very nice and patient when answering questions! Thank you so much! You should enter this website more often! :-)

Answer 23

More people like you is needed here!

Answer 24

Haha, thanks! I'm glad I can help!

Answer 25

\[\LARGE 2.\]The problem asks what the speed would be after 4.8 s \(\it{after}\) the bolt leaves the crossbow.

Now this is important; there are no forces on the bolt. Forces are what cause acceleration, which is the change in velocity. No force means no change in velocity.

Answer 26

\[\LARGE 3.\]Again, it seems that we're dealing with the \(4.8\ s\) after the bolt left the crossbow. It will have a speed of \(108\ m/s\).

You could think about this. If we know the bolt's speed (how many meters the bolt goes per second) and a time period (a number of seconds), then we can see how many meters it went during those seconds.

Average velocity is the change in distance per change in time.\[\bar v=\frac{\Delta d}{\Delta t}\]That's the definition for the math. The bar over the \(v\) in \(\bar v\) means we're using the average. I showed you that so maybe you recognize it another time - in case you need it. You want to know the distance change, \(\Delta d\).

Multiply both sides by \(\Delta t\).
I suggest you give it a try on your own, but here's how I worked through it. I hope you'll make sense of it!
\[\bar v=\frac{\Delta d}{\Delta t}\]
Multiply both sides by \(\Delta t\):
\[\bar v\ \Delta t=\frac{\Delta d\ \Delta t}{\Delta t}=\frac{\Delta d\ \cancel{\Delta t}}{\cancel{\Delta t}}=\Delta d\]Now you have\[\Delta d=\bar v\ \Delta t\]You know \(\bar v=108\ m/s\), and \(\Delta t =4.8\ s\).

Answer 27

If you have any more questions on this one, I'll see if you respond to this. Good luck!

Answer 28

Hey! I'm back! For problem #2, do I have to use the same formula I used for #1?

Answer 29

Who says you need a formula? :P

Answer 30

Tricky question. Sorry!

Answer 31

Haha! See? So do you understand #2?

Answer 32

Not really!

Answer 33

Wait! Since there are no forces on the bolt, and forces are what cause acceleration, which is the change in velocity. No force means no change in velocity, the answer will be 141 m/s^2, which is the initial acceleration of the bolt.

Answer 34

141 m/s^2, the initial velocity given in the problem.

Answer 35

Haha, the initial velocity leaving the bow is different. That was the acceleration. It is in the problem though!

Answer 36

I have to go. Good luck! I'll be back later!

Answer 37

So, is the answer 108 m/s?

Answer 38

For #2, you mean? Final answer?

Answer 39

Yes.

Answer 40

Sorry, I have to go, but I'll see if you respond to this. Thank you so much in advance...!

Answer 41

Haha, yep! You're right! Congrats! Take care!

Answer 42

So, the final answer for #2 is 108 m/s, right?

Answer 43

Yep! That's what we think!