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paulb203

Member

Registered: 2023-02-24

Posts: 337

Instantaneous acceleration from velocity time graph

Question 4.

Q. Work out the average acceleration during the 50 seconds?
I see that you can find this out by drawing a tangent to the graph, and working out the gradient of the line. And you can check this by a=delta v/(t)

Q. Estimate the time during the 50 seconds when the instantaneous acceleration = the average acceleration?
So, when does the inst.acc = 0.6m/s/s?
How do you find this out?

---

Prioritise. Persevere. No pain, no gain.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,637

Re: Instantaneous acceleration from velocity time graph

hi paulb203

Q. Work out the average acceleration during the 50 seconds?
I see that you can find this out by drawing a tangent to the graph, and working out the gradient of the line. And you can check this by a=delta v/(t)

Q. Estimate the time during the 50 seconds when the instantaneous acceleration = the average acceleration?
So, when does the inst.acc = 0.6m/s/s?
How do you find this out?

I'm not seeing an image of a graph. So I've made one up that has, I hope, similar features.

This is typical for a vehicle doing a drag race.  Acceleration is maximal at the start; but as the frictional forces build up it becomes harder and harder for the vehicle at add any more velocity so the acceleration tails off even though the velocity is still going up. Eventually the power in equals the power lost to friction and a maximum speed is reached.

My object starts from rest at A and accelerates as it travels towards B. The gradient curves down showing that the acceleration is not constant over time, but rather, that it is diminishing as the object approaches B.

The average acceleration is the (gain in velocity) over the (time taken) so the gradient of the line AB.  You won't need a tangent for this; just do (up) divided by (across).

I've drawn a line parallel to AB that touches the curve at C.  This is the instant when the gradient of the curve is equal to the gradient of AB.  So you need the time coordinate at C.

When is the instant acceleration 0.6 ?  I would draw any convenient line on the graph that has a gradient of 0.6.  Then I'd construct a parallel line that just touches the curve. That's the point where the acceleration is 0.6.

Hope that helps.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Instantaneous acceleration from velocity time graph

hi paulb203

Q. Work out the average acceleration during the 50 seconds?
I see that you can find this out by drawing a tangent to the graph, and working out the gradient of the line. And you can check this by a=delta v/(t)

Q. Estimate the time during the 50 seconds when the instantaneous acceleration = the average acceleration?
So, when does the inst.acc = 0.6m/s/s?
How do you find this out?

I'm not seeing an image of a graph. So I've made one up that has, I hope, similar features.

https://i.imgur.com/PDAykuV.gif

This is typical for a vehicle doing a drag race.  Acceleration is maximal at the start; but as the frictional forces build up it becomes harder and harder for the vehicle at add any more velocity so the acceleration tails off even though the velocity is still going up. Eventually the power in equals the power lost to friction and a maximum speed is reached.

My object starts from rest at A and accelerates as it travels towards B. The gradient curves down showing that the acceleration is not constant over time, but rather, that it is diminishing as the object approaches B.

The average acceleration is the (gain in velocity) over the (time taken) so the gradient of the line AB.  You won't need a tangent for this; just do (up) divided by (across).

I've drawn a line parallel to AB that touches the curve at C.  This is the instant when the gradient of the curve is equal to the gradient of AB.  So you need the time coordinate at C.

When is the instant acceleration 0.6 ?  I would draw any convenient line on the graph that has a gradient of 0.6.  Then I'd construct a parallel line that just touches the curve. That's the point where the acceleration is 0.6.

Hope that helps.

Bob

Bob,

I want you to do the same with my questions when needed. I mean, provide a graph.

paulb203

Member

Registered: 2023-02-24

Posts: 337

Re: Instantaneous acceleration from velocity time graph

Thanks a lot, Bob.

And, Doh! I forgot to post the link to the question (and graph);

https://www.mathsgenie.co.uk/resources/9-velocity-time-graphs.pdf

---

Prioritise. Persevere. No pain, no gain.

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Instantaneous acceleration from velocity time graph

hi paulb203

Q. Work out the average acceleration during the 50 seconds?
I see that you can find this out by drawing a tangent to the graph, and working out the gradient of the line. And you can check this by a=delta v/(t)

Q. Estimate the time during the 50 seconds when the instantaneous acceleration = the average acceleration?
So, when does the inst.acc = 0.6m/s/s?
How do you find this out?

I'm not seeing an image of a graph. So I've made one up that has, I hope, similar features.

https://i.imgur.com/PDAykuV.gif

This is typical for a vehicle doing a drag race.  Acceleration is maximal at the start; but as the frictional forces build up it becomes harder and harder for the vehicle at add any more velocity so the acceleration tails off even though the velocity is still going up. Eventually the power in equals the power lost to friction and a maximum speed is reached.

My object starts from rest at A and accelerates as it travels towards B. The gradient curves down showing that the acceleration is not constant over time, but rather, that it is diminishing as the object approaches B.

The average acceleration is the (gain in velocity) over the (time taken) so the gradient of the line AB.  You won't need a tangent for this; just do (up) divided by (across).

I've drawn a line parallel to AB that touches the curve at C.  This is the instant when the gradient of the curve is equal to the gradient of AB.  So you need the time coordinate at C.

When is the instant acceleration 0.6 ?  I would draw any convenient line on the graph that has a gradient of 0.6.  Then I'd construct a parallel line that just touches the curve. That's the point where the acceleration is 0.6.

Hope that helps.

Bob

Is this calculus?