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Need desperate help!

  • #1
I'm in big trouble, I have a Chapter test in Physics tomorrow & I don't get anything that will be on it. I can't answer any of the problems on the hw...I just don't get it. How do people find out the answers to these problems? I'm not one of those people who can spend 1 hour working on a single problem to find the answer, I don't have the time or the patience. Can someone PLEASE tell me that I don't have to do that to actually understand Physics?

One of my hw problems:

A motorcycle has a constant acceleration of 2.5 m/s2. Both the velocity and acceleration of the motocycle point in the same direction. How much time is required for the motorcycle to change its speed from (a) 21 to 31 m/s, and (b) 51 to 61 m/s?

Could somebody please help me work through this problem? I really just don't. know. what. to. do.

:( gwyneth
 

Answers and Replies

  • #2
Integral
Staff Emeritus
Science Advisor
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gwynethrosie said:
I'm in big trouble, I have a Chapter test in Physics tomorrow & I don't get anything that will be on it. I can't answer any of the problems on the hw...I just don't get it. How do people find out the answers to these problems? I'm not one of those people who can spend 1 hour working on a single problem to find the answer, I don't have the time or the patience.
You are in big trouble.

Can someone PLEASE tell me that I don't have to do that to actually understand Physics?
It is absolutely the best way to learn and understand physics. It is called.. effort.
One of my hw problems:

A motorcycle has a constant acceleration of 2.5 m/s2. Both the velocity and acceleration of the motocycle point in the same direction. How much time is required for the motorcycle to change its speed from (a) 21 to 31 m/s, and (b) 51 to 61 m/s?

Could somebody please help me work through this problem? I really just don't. know. what. to. do.

:( gwyneth
look at the units of acceleration.

[tex] \frac m {s^2} = \frac m s * \frac 1 s[/tex]

So the acceleration defines the change in velocity in one second. You are given 2 different velocities, how much does the velocity change? How do you compute that change? The acceleration tells you the rate of change, How could you get the time required?

BTW: If you have come here in hopes of finding someone to do your homework, you are going to be disappointed.
 
  • #3
Diane_
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Gwyneth - As Integral says, acceleration tells you how much the velocity changes per unit time. In this case, how many m/s it changes every second. In case (a) and case (b), both, you are asked how much time it would take for a 10 m/s change in the velocity.

Try this: You are making $2.50/hour. How many hours would you have to work to make $10?

There is a connection.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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The most important thing you can do is learn the definitions! Don't just say to yourself "Oh, yeah, I get that". MEMORIZE the precise words of the definitions.

When you saw the word "acceleration" you should have immediately thought "Okay, acceleration is defined as "change in speed divided by change in time". Then when you saw "change its speed from (a) 21 to 31 m/s" you should think "Okay, the speed changed from 21 to 31 so thats a change of 31-21= 10 m/s. Since the acceleration is 25 m/s2, it must be true that 25= 10/change in time. You can solve that equation to find the actual "change in time", the time it takes to change velocity by that much.

Yes, Integral's suggestion that you look at the units is very good but I recommend that you learn the definitions to start with.
 
  • #5
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You have to remember this equation [tex] a = \frac{Vf - Vo}{\Delta t} [/tex] where Vf = final speed, Vo = initial speed, a = acceleration, [tex] \Delta t [/tex] = time variation. Maybe Vf and Vo are Sf and So, I don't know.
 
  • #6
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LinkMage said:
You have to remember this equation [tex] a = \frac{Vf - Vo}{\Delta t} [/tex] where Vf = final speed, Vo = initial speed, a = acceleration, [tex] \Delta t [/tex] = time variation. Maybe Vf and Vo are Sf and So, I don't know.
well, in my school people deal only with differential, not with delta, so we come up only with differential euqations that the students don't know how to solve, so the teachers do it all then give the students the results
 

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