# Determine the acceleration

• Marioqwe
In summary, the problem involves finding the acceleration at t = 0 using the given velocity equation of v = 7.5(1 - 0.04x)^0.3, where v is in mi/h and x is in mi. The attempt at a solution uses the equation a = dv/dt and solves for a = -2.75 * 10^-4 ft/s^2. However, according to the textbook, the correct answer is -2.75 * 10^-6 ft/s^2.

## Homework Statement

The velocity is given by the relation v = 7.5(1 - 0.04x)^0.3 and v is in mi/h and x in mi. Determine the acceleration in ft/s^2 at t = 0 knowing that x = 0 when t = 0.

## The Attempt at a Solution

At t = 0, v = 7.5 mi/h, x=0
I know that a = v dv/dx

a = v[d(7.5(1 - 0.04x)^0.3)/dx] = (7.5)(7.5)(0.3)(-0.04)(1 - 0.04x)^-0.7

a = -0.675 mi/h * 5280ft/mi * h/3600 * h/3600 = -2.75 * 10^-4 ft/s^2

What am I doing wrong?

Last edited:
Marioqwe said:

## Homework Statement

The velocity is given by the relation v = 7.5(1 - 0.04x)^0.3 and v is in mi/h and x in h. Determine the acceleration in ft/s^2 at t = 0 knowing that x = 0 when t = 0.

## The Attempt at a Solution

At t = 0, v = 7.5 mi/h, x=0
I know that a = v dv/dx
That's incorrect. a = dv/dt, not v*dv/dt. (For some reason you are using x for time. t is nearly always used to represent time.)
Marioqwe said:
a = v[d(7.5(1 - 0.04x)^0.3)/dx] = (7.5)(7.5)(0.3)(-0.04)(1 - 0.04x)^-0.7

a = -0.675 mi/h * 5280ft/mi * h/3600 * h/3600 = -2.75 * 10^-4 ft/s^2

What am I doing wrong?

Mark44 said:
That's incorrect. a = dv/dt, not v*dv/dt. (For some reason you are using x for time. t is nearly always used to represent time.)

a IS equal to v*dv/dx=(dx/dt)*(dv/dx)=dv/dt. And I think x is supposed to be distance. In spite of the op saying the 'x is in h'. Otherwise what's the point of saying x=0 at t=0? Actually the final answer looks ok to me. There are a few typos in between.

Dick is right; x is supposed to be in mi (miles). I'm sorry for the confusion mark44. However, according to my textbook the answer is -2.75 * 10^-6 ft/s^2.

Dick said:
a IS equal to v*dv/dx=(dx/dt)*(dv/dx)=dv/dt.
Right. What threw me off was the OP saying that x was in units of hours.
Dick said:
And I think x is supposed to be distance. In spite of the op saying the 'x is in h'. Otherwise what's the point of saying x=0 at t=0? Actually the final answer looks ok to me. There are a few typos in between.

Marioqwe said:
Dick is right; x is supposed to be in mi (miles). I'm sorry for the confusion mark44. However, according to my textbook the answer is -2.75 * 10^-6 ft/s^2.

Well, I just checked it again and I still get -2.75*10^(-4) ft/s^2. Typo in the problem or the solution?

No typo in the problem so it must be the solution. Thanks for your help.

## What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It measures how quickly an object's speed is changing.

## How is acceleration measured?

Acceleration is typically measured in meters per second squared (m/s²) or feet per second squared (ft/s²).

## What factors affect acceleration?

The factors that affect acceleration include the mass of the object, the force acting on the object, and the direction of the force relative to the object's motion.

## How do you calculate acceleration?

Acceleration can be calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (vf - vi) / (tf - ti), where vf is the final velocity, vi is the initial velocity, tf is the final time, and ti is the initial time.

## What is the difference between positive and negative acceleration?

Positive acceleration occurs when an object speeds up, while negative acceleration (also known as deceleration) occurs when an object slows down. Negative acceleration can also be referred to as a change in direction, such as when an object changes from moving forward to moving backward.