How Do You Calculate Acceleration at Different Times from a Velocity-Time Graph?

[JadeLove]

A car travels at a constant velocity of +20 m/s for 5 seconds, than constantly accelerates up to + 40 m/s in 4 seconds. Then the car decelerates constantly until it reaches a velocity of 0 m/s, and rests for 1 second.

What is ax at t = 11 s? And what is ax at t = 3 seconds?

I'm sure if anyone could help me with either I would learn the process and be able to do the other by myself, thank you :)

 

[LawrenceC]

Isn't there a format for posing problems for which you need help?

 

[JadeLove]

Isn't there a format for posing problems for which you need help?

Dunno, I just joined like an hour ago... chill xD

- & I figured out how to solve this :)

 

[Cipherflak]

what do you mean with ax? acceleration times position? or is it street lingo for acceleration?

kids these days..

 

[rwisz]

I would approach this question by first analyzing the given information and understanding the concepts of velocity and acceleration.

From the given information, we can see that the car initially travels at a constant velocity of +20 m/s for 5 seconds, which means that its acceleration during this time is 0 m/s^2. This is because acceleration is the rate of change of velocity, and if the velocity is constant, there is no change.

After 5 seconds, the car accelerates constantly up to +40 m/s in 4 seconds. This means that its acceleration during this time is (40 - 20)/4 = 5 m/s^2. This is because the change in velocity is 40 m/s (final velocity) minus 20 m/s (initial velocity), divided by the time of 4 seconds.

Next, the car decelerates constantly until it reaches a velocity of 0 m/s, and rests for 1 second. This means that its acceleration during this time is (0 - 40)/4 = -10 m/s^2. This is because the change in velocity is 0 m/s (final velocity) minus 40 m/s (initial velocity), divided by the time of 4 seconds.

Now, to find the acceleration at t = 11 seconds, we need to look at the total time and velocity changes from the given information. From t = 0 to t = 5 seconds, the car has a constant velocity of +20 m/s, so its acceleration is 0 m/s^2. From t = 5 to t = 9 seconds, the car accelerates from +20 m/s to +40 m/s with an acceleration of 5 m/s^2. From t = 9 to t = 10 seconds, the car decelerates from +40 m/s to 0 m/s with an acceleration of -10 m/s^2. From t = 10 to t = 11 seconds, the car is at rest with an acceleration of 0 m/s^2. Therefore, the total acceleration at t = 11 seconds is 0 + 5 - 10 + 0 = -5 m/s^2.

At t = 3 seconds, the car has a constant velocity of +20 m/s, so its acceleration is 0 m/s^2.

I hope this explanation helps you understand the process and apply it to find the acceleration at