# Determine the acceleration

## 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:

Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor

## 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.)
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?

Dick
Homework Helper
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.

Mark44
Mentor
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.
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