Acceleration problem

  • #1
265
0
A car starts from rest and accelerates 3 m/s^2 for 3 sec. then continues with a constant velocity for 60 sec. Then accelerates at -5m/s^2 for 3 sec. then continues at a constant velocity for 30 sec.

What is the distance the car traveled
what is the average speed
displacement
average velocity

part of the solution:

for first 3 sec traveled 9 m/s and traveled 27m

then how do you find the constant velocity that the car traveled for 60 sec?
is it still 9m/s? if it is then it traveled 540m

then -15m/s for 3sec. or -15m

then is the constant velocity still 9m/s? If it is then it traveled 270m


If all of this is correct then the car traveled 792m
average speed= 8.25m/s
displacement = 792m
average velocity= 8.25m/s
 

Answers and Replies

  • #2
265
0
Going 9m/s for 27m
9m/s for 540m
-15m/s for -45m
-6m/s for -180m

Distance = 792m
Displacement = 342m
average speed = total distance/time = 8.25 m/s
average velocity= 3.56 m/s


Is this right?
 
  • #3
454
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You have the sections where the car accelerates wrong. The car doesn't move with 9m/s
for the first 3 seconds, but only at the end of this interval.
 
  • #4
265
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I dont understand the problem with that.
Since d=vt= 9 * 3= 27m
or is it d=.5at^2 = .5*3*(3)^2=13.5m
 
  • #5
79
0
You should break this down into four sections, one for each time interval (which you have started to do), and each section sort of flows into the next section.

The first 3 seconds, the car starts from rest and accelerates 3 m/s/s. You have to find two pieces of information, how far did the car travel, and how fast was it going at the end of the 3 seconds. You have:

t = 3s
a = 3m/s/s
u (beginning velocity) = 0 ("starts from rest")
v (final velocity) = ?
x (distance traveled) = ?

You cannot use d = vt as d = 3 * 3 because the velocity is not 3, the acceleration is.

First determine the distance travelled using the kinematic formula:

[tex] \Delta x = u + \frac{1}{2} a t^2[/tex]

Then determine the final velocity using the kinematic formula:

[tex] v_f = v + at [/tex]

Once you have that, you can move on to the next part. Look for the key words and what information that gives you, such as 'starts from rest' means initial velocity = 0, and 'constant speed', what does that tell you about the acceleration?

It's best to do each part carefully, as in this problem the 2nd part depends upon the answers of the 1st and so on. And a big hint - there will be no negative distances or velocities as you have written in you attempted response.
 
  • #6
265
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ok for the first part the vf=9m/s and s=13.5m

is the next constant velocity the 9m/s for the 60sec so vf=9m/s and s=vt=540m
then for the next part s=.5(5)(3^2)=22.5m and vf=-6 or does s=-6*3=-18m
and then what is the next constant velocity -6m/s????
 
  • #7
79
0
Hmmm, your calculations look correct, are you sure the problem is copied correctly? To go from 9 m/s to -6 m/s would mean the car came to a screeching stop and then immediately went into reverse. Check to make sure the acceleration isn't something like
-.5m/s/s.
 
  • #8
265
0
It is -5m/s^2
vf=9m/s and s=13.5m
vf=9m/s and s=vt=540m
since constant acceleration s=.5at^2=.5(-5)(9)=-22.5m vf=-6m/s
and then what is the next constant velocity -6m/s s=vt=-6*30=-180m

displacement= 13.5+540+-22.5-180m= 351m
distance = 13.5+540+22.5+160=736m
avg speed= total distance/total time= 736/(60+3+3+30)=7.7m/s
avg velocity = displacement/time = 351/96=3.7m/s

Is this right?????
 
  • #9
79
0
I haven't double-checked all your calculations, (my calculator is done for the night) but what you're doing looks good to me.
 
  • #10
265
0
for the second change in velocity I forgot to add v0t to the s=v0t+.5at^2
so that s=-4.5m
so displacement= 13.5+540+-4.5+-180=369m
distance= 13.5+540+4.5+180=738
avg speed= 7.7m/s
avg velocity=3.8m/s

I think this is how u do it
 
  • #11
79
0
O.k., I pulled my calculator out, I think it should be a positive 4.5m, not negative:

s = 9*3 + .5*-5*9
= 27 + -22.5
= 4.5m (not -4.5m)

Double check it, I may have made a mistake.
 
  • #12
265
0
my bad

thanks
 

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