Find maximum velocity in Acceleration-Time Graph

In summary: So, in summary, the maximum velocity of the particle can be found by calculating the area under the acceleration-time graph, which is equal to the change in velocity over time. This can be done by using the formula a(t) = (110 - 10(t-1))/11, integrating it to find the velocity as a function of time, or by using the known theorem that the area under a triangle is equal to 1/2 of its base times its height. In this case, the maximum velocity is found to be 55m/s.
  • #1
Govind_Balaji
83
0

Homework Statement


Find the maximum velocity of the particle using the graph given.(See attachment)

Homework Equations



Initial velocity was not given. I assumed it to be 0.

In book, the solution was 55m/s^2

The Attempt at a Solution



Initial acceleration was ##10m/s^2##. Final acceleration after 11 seconds was ##0m/s^2##.

I figured out that ##a_n=\frac{110-10(n-1)}{11}##, where n is the second and ##a_n## is the acceleration in ##n##th second.

I hope I am right at this step.

I think that velocity will be maximum at 11th second.

So I calculate the velocity at 1st second by
[itex]\begin{align*}v_n&=v_{n-1}+\frac{a_n}{2}\\
&=v_{n-1}+\frac{55-5(n-1)}{11}\\\end{align*}
[/itex].

I derived this using identities of motion and my already derived ##a_n## formula.

[itex]
v_0=0\\
v_1=0+\frac{55-5(1-1)}{11}=5\\
v_2=5+\frac{55-5(2-1)}{11}=\frac{105}{11}\\
...\\
v_{11}=295/11\\
[/itex]

I should get 55m/s.

Please help me out.
 

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  • #2
Govind_Balaji said:
I figured out that ##a_n=\frac{110-10(n-1)}{11}##, where n is the second and ##a_n## is the acceleration in ##n##th second.
Does this equation recover the values of a at n=0 and n=11 that you know from looking at the graph?
 
Last edited:
  • #3
No, you can't do that. Don't use n. Use t and find an expression for the acceleration as a function of time a(t) where t is any number - not just a whole number. Than integrate the acceleration to find the velocity. If you don't know integrals than you probably know some theorem relating the change in velocity to the area in a graph of acceleration vs time. In this case the area is a triangle which can be solved without integrals.
 
  • #4
dauto said:
No, you can't do that. Don't use n. Use t and find an expression for the acceleration as a function of time a(t) where t is any number - not just a whole number. Than integrate the acceleration to find the velocity. If you don't know integrals than you probably know some theorem relating the change in velocity to the area in a graph of acceleration vs time. In this case the area is a triangle which can be solved without integrals.

CAF123 said:
Does this equation recover the values of a at n=0 and n=11 that you know from looking at the graph?

Wait a minute!

I got a brilliant idea.

I know that the area below velocity-time graph gives the total displacement of the body.

I thought about it a while.

Velocity is the change of displacement over time.

Acceleration is the change of velocity over time.

Then , presto! Instead of velocity in velocity-time graph, we have acceleration.

Instead of distance(factor of velocity), we have velocity(factor of acceleration).

Hence the area under the graph is the max. velocity i.e. 55m/s.

After solving it, I came to this thread and also saw dauto saying the same.

Thank you CAF123.
 
  • #5
Govind_Balaji said:
Wait a minute!

I got a brilliant idea.

I know that the area below velocity-time graph gives the total displacement of the body.

I thought about it a while.

Velocity is the change of displacement over time.

Acceleration is the change of velocity over time.

Then , presto! Instead of velocity in velocity-time graph, we have acceleration.

Instead of distance(factor of velocity), we have velocity(factor of acceleration).

Hence the area under the graph is the max. velocity i.e. 55m/s.

After solving it, I came to this thread and also saw dauto saying the same.

Thank you CAF123.


Excellent.
 

1. What is an acceleration-time graph?

An acceleration-time graph is a graphical representation of an object's acceleration over a specific period of time. The acceleration is usually plotted on the y-axis, while time is plotted on the x-axis.

2. How can I find the maximum velocity on an acceleration-time graph?

The maximum velocity on an acceleration-time graph can be found by identifying the point on the graph where the slope or rate of change is the steepest. This point represents the highest velocity reached by the object.

3. What does the slope of an acceleration-time graph represent?

The slope of an acceleration-time graph represents the rate of change of acceleration over time. A steeper slope indicates a higher rate of acceleration, while a flatter slope indicates a slower rate of acceleration.

4. How do I calculate the maximum velocity from an acceleration-time graph?

To calculate the maximum velocity from an acceleration-time graph, you can use 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. Plug in the values for acceleration and time at the point where the slope is the steepest to find the maximum velocity.

5. Can the maximum velocity on an acceleration-time graph be negative?

Yes, the maximum velocity on an acceleration-time graph can be negative. This indicates that the object is moving in the opposite direction of its initial velocity. The magnitude of the negative velocity is still the maximum velocity reached by the object.

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