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Instantaneous Velocity and Acceleration

  • Thread starter VaioZ
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  • #1
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Homework Statement


A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following
a. The average velocity of the car from t=0s to t=10s
b. The instantaneous velocity of the car at t=10s
c. The average acceleration of the car at t=10s
d. The instantaneous acceleration of the car at t=10.s

Homework Equations


Derivatives

The Attempt at a Solution


in a.) I integrated the equation to get the position function
x(t)=(1/2)(7.20m/s^2)(t^2)+(1/3)(-0.720m/s^3)(t^3)+C
then I substitute t = 10 s
I got 120 m/s but the answer should be -72 m/s <- So I need help here


b.) I derive the x(t) to return the original equation then I substitute t = 10 s right?
c.) I'll simply derive the original equation (7.20m/s^2)+2(-0.720)(10) = -7.2 m/s^2
d.) Use the original equation then plug in t = 10 s ? <- Help
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following
a. The average velocity of the car from t=0s to t=10s
b. The instantaneous velocity of the car at t=10s
c. The average acceleration of the car at t=10s
d. The instantaneous acceleration of the car at t=10.s

Homework Equations


Derivatives

The Attempt at a Solution


in a.) I integrated the equation to get the position function
x(t)=(1/2)(7.20m/s^2)(t^2)+(1/3)(-0.720m/s^3)(t^3)+C
then I substitute t = 10 s
I got 120 m/s but the answer should be -72 m/s <- So I need help here
If x(t) is the position of the car, why are the units m/s? That's your first problem.

Exactly which question does -72 m/s answer?
 
  • #3
20,209
4,253

Homework Statement


A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following
a. The average velocity of the car from t=0s to t=10s
b. The instantaneous velocity of the car at t=10s
c. The average acceleration of the car at t=10s
d. The instantaneous acceleration of the car at t=10.s

Homework Equations


Derivatives

The Attempt at a Solution


in a.) I integrated the equation to get the position function
x(t)=(1/2)(7.20m/s^2)(t^2)+(1/3)(-0.720m/s^3)(t^3)+C
then I substitute t = 10 s
I got 120 m/s but the answer should be -72 m/s <- So I need help here
Where did the -72 m/s come from? I calculated that the distance covered over 10 sec was 120 m. So the average velocity should be 12 m/s. During the 10 sec, the instantaneous velocity is never zero, so the average certainly shouldn't be negative.

Chet
 
  • #4
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Where did the -72 m/s come from? I calculated that the distance covered over 10 sec was 120 m. So the average velocity should be 12 m/s. During the 10 sec, the instantaneous velocity is never zero, so the average certainly shouldn't be negative.

Chet
I borrowed my friend's notebook because I studying about kinematic then I saw this example and resolved it. The answer on a letter A is -72(on his notebook) but I got 120 soooooo I posted in this forum to confirm which is the right answer
 
  • #5
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If x(t) is the position of the car, why are the units m/s? That's your first problem.

Exactly which question does -72 m/s answer?
oooops I meant to say 120 m
 
  • #6
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4,253
I borrowed my friend's notebook because I studying about kinematic then I saw this example and resolved it. The answer on a letter A is -72(on his notebook) but I got 120 soooooo I posted in this forum to confirm which is the right answer
So, from what you know now, was his answer right or not?
 
  • #7
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So, from what you know now, was his answer right or not?
First I was frustrated that I can't get -72 m which consuming all my time so I decided to post my silly question (sorry guys I have to do it time is gold right?)

but I conclude that maybe he miscalculate his problem or something
but it is confirm the answer in a.) is 120 m? I always get 120 m. Sooooooo I hope I'm right. This problem is already wasted 1 hour in my review session.
 
  • #8
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4,253
First I was frustrated that I can't get -72 m which consuming all my time so I decided to post my silly question (sorry guys I have to do it time is gold right?)

but I conclude that maybe he miscalculate his problem or something
but it is confirm the answer in a.) is 120 m? I always get 120 m. Sooooooo I hope I'm right. This problem is already wasted 1 hour in my review session.
What is the definition of average velocity? Does velocity have units of meters?

Chet
 
  • #9
19
0
What is the definition of average velocity? Does velocity have units of meters?

Chet
I meant to say when you're getting x(t) the units should m or meters right? So I answered 120 m. Then I use average velocity = ΔX/ΔT soooo 120 m / 10 s = 12 m/s right?
This is for letter A
 
  • #10
20,209
4,253
I meant to say when you're getting x(t) the units should m or meters right? So I answered 120 m. Then I use average velocity = ΔX/ΔT soooo 120 m / 10 s = 12 m/s right?
This is for letter A
Correct.

Chet
 
  • #11
19
0
Correct.

Chet
for letter B finding the in instantaneous velocity. I don't need to derive it because it is already in v(t)
I'll just plug t = 10 s into v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2

I got 0 m/s hmmm I find it odd to get 0 in instantaneous velocity
 
  • #12
20,209
4,253
for letter B finding the in instantaneous velocity. I don't need to derive it because it is already in v(t)
I'll just plug t = 10 s into v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2

I got 0 m/s hmmm I find it odd to get 0 in instantaneous velocity
Have you ever heard of speeding up, then slowing down, and finally stopping.

Chet
 
  • #13
19
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Have you ever heard of speeding up, then slowing down, and finally stopping.

Chet
Oh okay that make sense to prove that it I can get 0 m/s thank you!!!
 
  • #14
107
34
Very good job. You were very close on a.
..The integration was correct.
..Just a minor math error

Homework Statement


A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following
a. The average velocity of the car from t=0s to t=10s

The Attempt at a Solution


in a.) I integrated the equation to get the position function
x(t)=(1/2)(7.20m/s^2)(t^2)+(1/3)(-0.720m/s^3)(t^3)+C
then I substitute t = 10 s
I got 120 m/s but the answer should be -72 m/s <- So I need help here
1) Remember, you are calculating average velocity for the 10 second period. Since the initial velocity is 0 m/s. The final velocity is 0 m/s. And all velocities in between are greater than zero there is no way the answer can be a negative number.

2) You have to dispose of the C by applying the equation when t=0. Naturally C= 0 so you can dispense with it.

3) You calculated x(t) at 10 seconds to be 120 Meters (not m/s). So you got the distance traveled in ten seconds correct.

4) You are going for average velocity. so that is the distance traveled divided by the time. v avg = [x(10s) - x (0s) ] / [t(10s) - t(0s)]
V avg = 120 m / 10 s = 12 m/s

Question: Now how do you check the answer?
 
  • #15
107
34
What is the difference between AVERAGE acceleration and INSTANTANEOUS acceleration.
{Perhaps c. should be from t=0s to t=10s}

c. The average acceleration of the car at t=10s
d. The instantaneous acceleration of the car at t=10.s
 
  • #16
107
34
Practicing Latex

Velocity
[tex] v(t)=(7.20m/s^2) \times t-(0.720m/s^3) \times t^2[/tex]

Distance traveled is the area under the curve.
So he needed to integrate the velocity equation, which he did correctly.
[tex] \int v(t) \, dt = x(t) = \frac {7.20m/s^2 \times t^2} {2} \quad + \quad \frac {-0.720m/s^3 \times t^3} {3} \quad+ \quad C [/tex]

And Acceleration is the slope of the velocity line.
So he needed to derive the velocity equation, which he did correctly.
[tex] \frac {\partial v(t)} {\partial t}= a(t) = 7.20m/s^2 \quad + \quad 2 \times (-0.720)m/s^3 \times t \quad [/tex]
 

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