# Calculate Acceleration of Automobile at t = 4s

• davev
In summary: It's because acceleration is the slope of your velocity curve. It's not v/t. It's dv/dt, the derivative or approximately Δv/Δt if you pick two points around your point. It's the rate of change of velocity, not v/t.

#### davev

An automobile moves forward and backward on a straight highway.The graph shows the velocity of this automobile as a function of time.

What is the acceleration of the automobile at t = 4 s? (Round to two significant digits.)

a = change in velocity / time

I said the acceleration was zero, because the automobile would be stopped at t = 4 s. This is wrong, however.

davev said:
An automobile moves forward and backward on a straight highway.The graph shows the velocity of this automobile as a function of time.

What is the acceleration of the automobile at t = 4 s? (Round to two significant digits.)

a = change in velocity / time
I said the acceleration was zero, because the automobile would be stopped at t = 4 s. This is wrong, however.

Of course it is wrong. The automobile may be stopped, but acceleration is the rate of change of velocity. It's just in between moving forwards and moving backwards. Try that again.

Dick said:
Of course it is wrong. The automobile may be stopped, but acceleration is the rate of change of velocity. It's just in between moving forwards and moving backwards. Try that again.

The question is asking for the acceleration at exactly 4 s, so if I used the formula a = v/t I would get a = 0/4, which is 0.

The answer is -10, but I don't understand why when that acceleration is reached a second later.

davev said:
The question is asking for the acceleration at exactly 4 s, so if I used the formula a = v/t I would get a = 0/4, which is 0.

The answer is -10, but I don't understand why when that acceleration is reached a second later.

It's because acceleration is the slope of your velocity curve. It's not v/t. It's dv/dt, the derivative or approximately Δv/Δt if you pick two points around your point. It's the rate of change of velocity, not v/t. As you almost said to begin with it's (change of velocity)/(change of time).

Last edited:
The acceleration would be the slope of the tangent line at t = 4 s.

I would first clarify the units of the velocity graph, as it is not specified in the given information. Assuming the units are in meters per second (m/s), I would proceed with the calculation by using the formula for acceleration: a = (v2 - v1) / (t2 - t1), where v2 and v1 are the velocities at times t2 and t1 respectively.

In this case, at t = 4 s, the velocity is approximately 0 m/s, as the graph shows the automobile coming to a stop. However, since the graph is a curve, we cannot simply take the velocity at t = 4 s as the final velocity. Instead, we need to find the slope of the tangent line at t = 4 s, which represents the instantaneous velocity at that specific time.

To find the slope of the tangent line, we can use the concept of derivatives from calculus. The derivative of a function at a specific point represents the slope of the tangent line at that point. Therefore, we can use the derivative of the velocity function at t = 4 s to calculate the acceleration at that point.

Without the specific form of the velocity function, I cannot provide an exact value for the acceleration. However, I can provide the general steps for calculating it. First, we need to find the derivative of the velocity function. Then, we can plug in t = 4 s into the derivative function to find the slope of the tangent line at that point, which represents the acceleration. Finally, we can round the result to two significant digits to provide the requested answer.

In summary, as a scientist, I would use the concept of derivatives to find the acceleration of the automobile at t = 4 s, taking into account the units of the velocity graph. This would provide a more accurate and precise calculation compared to simply assuming the acceleration is zero.

## What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a measure of how quickly an object's speed is increasing or decreasing.

## How do you calculate acceleration?

To calculate acceleration, you need to know the change in velocity and the time it takes for that change to occur. The formula for acceleration is: a = (vf - vi) / t, where a is acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

## What is the unit of measurement for acceleration?

The unit of measurement for acceleration is meters per second squared (m/s²).

## What information do you need to calculate the acceleration of an automobile at t = 4s?

To calculate the acceleration of an automobile at t = 4s, you will need to know the initial velocity, final velocity, and the time interval of 4 seconds. These values can be obtained from a speedometer, odometer, or by using a stopwatch to measure the time interval.

## Why is it important to calculate the acceleration of an automobile?

Calculating the acceleration of an automobile is important for understanding how the vehicle is moving and how it may be affected by external forces such as friction, air resistance, or gravity. It can also be used to determine the performance of the vehicle and identify any potential issues that may need to be addressed.