Kinematics problem with constant acceleration.

In summary: you are given d, and t, and you know v2, so solve for v1. then for part b) vf^2 = vi^2 + 2*a*d you know vf, and vi, and d. solve for a. for part c) you need to solve for the time it takes for the car to travel the first 56.1 m. the time for the first 56.1 m is the same time it takes for the last 56.1 m, because the acceleration is constant. so, figure out the time for the last 56.1 m, and then use that time to figure out how much distance it travelled before the first 56.
  • #1
motoxxx
4
0

Homework Statement


A car moving with a constant acceleration covered the distance between two points 56.1 m apart in 5.44s. Its speed as it passes the second point was 14.2m/s. (a) what was the speed at the first point? (b) What was the acceleration?. (c) At what prior distance from the first point was the car at rest?


Homework Equations



These are all I came up with. This one has me stumped.

vf = vi + a delta t ------ (a)

ax = v2x - v1x --------(b)
_________
t2 - t1

delta x = 1/2 (vf +vi ) delta t -------- (c)




The Attempt at a Solution



(a)vf = 56.1 + x(delta t)
=?

(b) assuming that acceleration is constant, then between the two known points would it be 10.31 m/s^2?

OR

v avg = 14.2 +0
_______
2
= 7.1 m/s

Therefore,

ax = vx * t
= 7.1 * 5.22
=39.22 m/s^2

(c) ------ No idea where to start.

Help!
 
Physics news on Phys.org
  • #2
Can anyone point me in the right direction?
 
  • #3
I can try to help you. Do you know any other kinematic equations? Of the equations you posted, b looks similar to a, just rearranged, and I don't know where the x came from, and c is totally foreign to me.

A couple of key points, velocity is changing and I don't think an average velocity does you any good.

I don't quite understand what you did. I've done the calculations and have what seems to be a correct answer (I put my answers back to test them). To begin with, you need vi and a, two unknowns, so you want to find two equations to use, then solve for those unknowns. Once that is done, you can do part c.

That's my suggestion, someone else may have a better idea.
 
  • #4
I don't even know how to derive the equations for this type of question. Is there any way you can show an example of what you did to get me started?

thanks
 
  • #5
They are:

[tex]\Delta x = v_1t + \frac{1}{2}at^2[/tex]

[tex]vf = vi + at[/tex]
[tex]vf^2 = vi^2 + 2a \Delta x[/tex]

The second one is the same as your 'a' equation. You can see that you don't have quite enough information to complete any of them, but if you start filling stuff in, you should be able to get to a point where you have two equations with two unknowns. This will get you a and b of your problem. Then, to solve for c, for now I'm going to set you off on your own. If you run into more trouble I'll help.
 
  • #6
Man, since you wrote that reply, I have been plugging in numbers every which way and I I am about ready to throw my textbook out the god damn window!

I have substituted equations for the unkown's left right and center and gotten nowhere.

I have used the equations you provided and have not been successful at all.

For the love of god, show me the answers!
 
  • #7
motoxxx said:
Man, since you wrote that reply, I have been plugging in numbers every which way and I I am about ready to throw my textbook out the god damn window!

I have substituted equations for the unkown's left right and center and gotten nowhere.

I have used the equations you provided and have not been successful at all.

For the love of god, show me the answers!

for part a)

d = [(v1+v2)/2]*t

solve for v1.

there are different ways to do the problems, but this one is probably the most direct way.
 

FAQ: Kinematics problem with constant acceleration.

What is kinematics?

Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion.

What is constant acceleration?

Constant acceleration is when an object's velocity changes at a constant rate over a period of time. This means that the object's speed increases or decreases by the same amount in each unit of time.

How do you calculate velocity in a kinematics problem with constant acceleration?

You can calculate velocity by using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time interval.

What is the difference between speed and velocity in kinematics?

Speed is a measure of how fast an object is moving, while velocity is a measure of both the speed and direction of an object's motion. In other words, velocity takes into account both the magnitude and direction of an object's velocity, while speed only considers the magnitude.

How do you solve a kinematics problem with constant acceleration?

To solve a kinematics problem with constant acceleration, you can use the four main kinematic equations: v = u + at, s = ut + 1/2at^2, v^2 = u^2 + 2as, and s = (u + v)t/2. These equations involve variables such as velocity, initial velocity, acceleration, time, and displacement, and can be rearranged to solve for any missing variable.

Similar threads

Replies
5
Views
2K
Replies
11
Views
940
Replies
9
Views
2K
Replies
3
Views
15K
Replies
2
Views
645
Replies
6
Views
1K
Replies
2
Views
4K
Back
Top