# Instantaneous acceleration from Velocity-time graph

• James2911
In summary, the conversation is about finding the weight of a person standing on a lift as it moves vertically upwards, given a v-t graph. The person calculates the instantaneous acceleration by drawing a tangent at a specific point on the graph and using the formula v2 - v1/t2 - t1 = a. They use this acceleration value to solve for the weight using the equation N = m(g+a). They also discuss the difference between average and instantaneous acceleration and how using points on a tangent can give different results depending on the time interval. The problem statement is given as well.
James2911

## Homework Statement

https://imgur.com/a/bvcxjNy

## Homework Equations

Since the elevator is accelerating upwards,
N - mg = ma
or N = m(g+a) ------------ (1)

## The Attempt at a Solution

In order to get the accelerations at that instant, I drew a tangent at that point and used v2 - v1/t2 - t1 = a
and got;
for (i) t = 1s
v2 - v1/t2 - t1 = a;
20-10/2-1 = a
10 m/s^2 = a
Applying this to equation (1)
N = m(10+10) = 20m

(ii) Following the same steps as in (i)
20-20/8-2 = 0 m/s^2
N = m(10+0) = 10m

(iii) 10-20/12-10 = -10/5
a = -5m/s^2
N = m(10-5) = 5m

The answer matches the solution but I have a doubt about the process of arriving at instantaneous acceleration.
The tangent gives the instantaneous acceleration and we can use any points on the tangent to calculate to the acceleration.So for eg,

In (ii) above I used; v2 = 20, v1 = 20, t2 = 8 and t1 = 2 secs
But if instead of those I took v2 = 20, v1 = 0 t2 = 8 and t1 = 0 why does my answer comes out wrong?

Please place parentheses where they belong. The expression (v2-v1)/(t2-t1) gives the average acceleration over the time interval t1 to t2. When the slope is constant (straight line) the average acceleration is equal to the instantaneous acceleration. The slope is constant from 0 to 2 s, from 2 to 10 s and from 10 to 14 s. It is not constant from 0 to 8 s. Over that time interval the acceleration is changing so the expression gives the average and not the instantaneous value.

Please type the problem statement and post a picture, not a link (that's gone soon)

James2911 said:
The tangent gives the instantaneous acceleration... But if instead of those I took v2 = 20, v1 = 0 t2 = 8 and t1 = 0 why does my answer comes out wrong?

Because those aren't two points on the same tangent. The line between those points cuts across the graph, it isn't tangent to it.

Since this graph consists of straight lines, the tangent in each segment is just a line coinciding with that straight line.

kuruman said:
Please place parentheses where they belong. The expression (v2-v1)/(t2-t1) gives the average acceleration over the time interval t1 to t2. When the slope is constant (straight line) the average acceleration is equal to the instantaneous acceleration. The slope is constant from 0 to 2 s, from 2 to 10 s and from 10 to 14 s. It is not constant from 0 to 8 s. Over that time interval the acceleration is changing so the expression gives the average and not the instantaneous value.
Thank you so much!

James2911 said:
Thank you so much!
As @BvU said, you should post any picture. Here's your's:

Also:
You should type the complete problem statement (even if it's given in the posted image or in the thread title).

A person of mass M (in kg) is standing on a lift. If the lift moves vertically upward, according to given v-t graph, then find out the weight of the man at the following instants: (g = 10m/s2).

(i) t = 1 seconds

(ii) t = 8 seconds

(iii) t = 12 seconds​

#### Attachments

• hfnebcO[1].jpg
17.5 KB · Views: 544

## 1. What is instantaneous acceleration?

Instantaneous acceleration is the rate of change of velocity at a specific moment in time. It represents how quickly an object's velocity is changing at a particular instant.

## 2. How is instantaneous acceleration calculated from a velocity-time graph?

Instantaneous acceleration can be calculated by finding the slope of the tangent line at a specific point on a velocity-time graph. This is equivalent to finding the derivative of the velocity function with respect to time.

## 3. What does a positive instantaneous acceleration on a velocity-time graph indicate?

A positive instantaneous acceleration on a velocity-time graph indicates that the object is speeding up. This means that its velocity is increasing over time.

## 4. What does a negative instantaneous acceleration on a velocity-time graph indicate?

A negative instantaneous acceleration on a velocity-time graph indicates that the object is slowing down. This means that its velocity is decreasing over time.

## 5. How does the slope of a velocity-time graph relate to instantaneous acceleration?

The slope of a velocity-time graph represents the rate of change of velocity, which is equivalent to instantaneous acceleration. This means that the steeper the slope, the greater the instantaneous acceleration, and vice versa.

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