Acceleration-Time Graph and Velocity Multiple Choice Question

In summary, the graph shows that the acceleration of an object is 1 m/s^2 from t=0 to t=2 and then increases by 1 m/s^2 every 2 seconds. The change in velocity of the object from t=0 to t=6 is 9 m/s, as calculated using both methods of areas and calculus. The given answer of 10 m/s is incorrect.
  • #1
AntSC
65
3
Homework Statement
Finding ##\Delta v## from an ##a-t## graph
Relevant Equations
Area under graph methods
The variation with time t of the acceleration a of an object is shown

1640081649828.png


What is the change in velocity of the object from ##t=0## to ##t=6##?

A. ##6ms^{-1}##
B. ##8ms^{-1}##
C. ##10ms^{-1}##
D. ##14ms^{-1}##

So apparently the answer is B, which I am having trouble reconciling.

Using methods of areas I get ##2ms^{-1}## for the first ##2s## and ##8ms^{-1}## for the last ##4s##. Total ##\Delta v=10ms^{-1}## right?
I don't think there should be much more calculation needed for this question so I assume my interpretation is what is off.

I've tried doing this with calculus and I get the same answer, which isn't a surprise. I even tried assuming that ##v=0## at the start and built it up from there, which still gives me a ##\Delta v## of ##10ms^{-1}##.

I'm a bit dumbfounded. Can anyone comment?
 
Physics news on Phys.org
  • #2
Hello,

Well, for what it's worth: I completely agree with you. :smile:
Here in PF we have often seen given answers to be mistaken and for now I assume that is the case here too !

##\ ##
 
  • Like
Likes jbriggs444, AntSC and vcsharp2003
  • #3
BvU said:
Hello,

Well, for what it's worth: I completely agree with you. :smile:
Here in PF we have often seen given answers to be mistaken and for now I assume that is the case here too !

##\ ##
Thanks BvU!
You've helped my sanity.
It's always uncertain with answers for multiple-choice questions as there's no working, so you can't retrace any steps.
Glad there's some agreement here
 
  • #4
Wait a minute. I am afraid this problem makes me question my sanity. The acceleration increases by 1 m/s2 every 2 s. If it is 1 m/s2 at ##t= 2## s, then it must be zero at ##t=0##. The change in velocity ##\Delta v## from 0 s to 6 s is the area of the right triangle of base 6 s and height 3 m/s2. This is 9 m/s.

Using calculus, ##a=\frac {1}{2}t##. Then $$\Delta v= \frac {1}{2}\int_0^6t~dt=\left. \frac{1}{4}t^2 \right|_0^6= \frac{1}{4}\times (36-0) = 9~\text{m/s}$$Why is the correct answer 10 m/s? I don't believe that the correct answer is one of the given choices.

On edit: My sanity is restored. All of the above is incorrect but I need new glasses. See post #8.
 
Last edited:
  • #5
kuruman said:
then it must be zero at t=0.
The graph clearly shows an acceleration of 1 m/s2 at t=0

##\ ##
 
  • #6
kuruman said:
Wait a minute. I am afraid this problem makes me question my sanity. The acceleration increases by 1 m/s2 every 2 s. If it is 1 m/s2 at ##t= 2## s, then it must be zero at ##t=0##. The change in velocity ##\Delta v## from 0 s to 6 s is the area of the right triangle of base 6 s and height 3 m/s2. This is 9 m/s.

Using calculus, ##a=\frac {1}{2}t##. Then $$\Delta v= \frac {1}{2}\int_0^6t~dt=\left. \frac{1}{4}t^2 \right|_0^6= \frac{1}{4}\times (36-0) = 9~\text{m/s}$$Why is the correct answer 10 m/s? I don't believe that the correct answer is one of the given choices.

But the acceleration function is a piecewise function and the ##\frac{1}{4}t^2## function you have used only applies in the ##1\leq t\leq 6## domain and not ##0\leq t\leq 6##. I did test this with integration too and still got ##10ms^{-1}## but maybe check again if you have the inclination.

I am not saying the answer is ##10ms^{-1}##, just that I don't see what else it could be. Perhaps we're all making errors in different places LOL

I'm sure we can crack this.

Also thought that I would add that this question is from an official international baccalaureate standard level exam paper. You would think the published answers would have been at least triple-checked but who knows.
 
  • #7
AntSC said:
I'm sure we can crack this.
You have already done so. Your answer is right and the answer key is wrong. It is an important lesson to learn that sometimes one can be right and the authorities can be wrong. [Just don't let it go to your head].

My guess is that someone wanted the graph to depict an acceleration of zero for the first two seconds, did the calculation that way and never checked to see whether the graph agreed with the answer.
 
  • Like
Likes BvU
  • #8
BvU said:
The graph clearly shows an acceleration of 1 m/s2 at t=0

##\ ##
Oops! My tired old eyes didn't discern that the horizontal line from {0,1} to {2,1} is not dashed like its vertical counterpart. I stand corrected.
 
  • #9
It took me several looks as well ! And that the vertical scale steps are 1 and the horizontal ones 2 didn't help much either :smile:

Oh well, keeps us sharp, I suppose :wink:

##\ ##
 
  • Like
Likes kuruman, hmmm27 and jbriggs444

Related to Acceleration-Time Graph and Velocity Multiple Choice Question

1. What is an acceleration-time graph?

An acceleration-time graph is a visual representation of an object's acceleration over time. It plots the change in velocity on the y-axis and the change in time on the x-axis. The slope of the graph represents the object's acceleration, with a steeper slope indicating a greater acceleration.

2. How do you interpret an acceleration-time graph?

To interpret an acceleration-time graph, you can look at the slope of the graph. A positive slope indicates a positive acceleration, meaning the object is speeding up. A negative slope indicates a negative acceleration, meaning the object is slowing down. A horizontal line indicates a constant velocity, and a vertical line indicates an instantaneous change in velocity.

3. What is the relationship between velocity and acceleration on an acceleration-time graph?

The relationship between velocity and acceleration on an acceleration-time graph is that acceleration is the rate of change of velocity. This means that the slope of the graph represents the object's acceleration, while the y-intercept represents the initial velocity of the object.

4. How can you calculate the acceleration from an acceleration-time graph?

To calculate the acceleration from an acceleration-time graph, you can use the formula a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. You can also find the slope of the graph by choosing two points on the graph and dividing the change in velocity by the change in time.

5. How is an acceleration-time graph different from a velocity-time graph?

An acceleration-time graph plots the change in velocity over time, while a velocity-time graph plots the actual velocity of an object over time. The slope on an acceleration-time graph represents acceleration, while the slope on a velocity-time graph represents the object's velocity. Additionally, the y-intercept on an acceleration-time graph represents initial velocity, while the y-intercept on a velocity-time graph represents displacement.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
971
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
3
Replies
98
Views
4K
  • Introductory Physics Homework Help
Replies
17
Views
2K
Back
Top