# Velocity v(t) = b(t-T)^2, find position function x(t)

• courtney1121
In summary, the correct function that describes x(t) is c. Taking the derivative of c gives (1/3)3b(t-T)^2, which simplifies to b(t-T)^2 and matches the given x-velocity function.
courtney1121
If a car's x-position at time t = 0 is x(0) = 0 and it has an x-velocity of
vx(t) = b(t-T)^2, where b and T are constants, which function best describes x(t)?

a. x(t) = 2b(t - T)
b. x(t) = 3b(t - T)3
c. x(t) = (1/3)b(t - T)3
d. x(t) = (1/2)b(t - T)
e. x(t) = (1/3)b[(t - T)3 + T3]
f. None of the above

I was thinking d, but not sure.

courtney1121 said:
If a car's x-position at time t = 0 is x(0) = 0 and it has an x-velocity of
vx(t) = b(t-T)^2, where b and T are constants, which function best describes x(t)?

a. x(t) = 2b(t - T)
b. x(t) = 3b(t - T)3
c. x(t) = (1/3)b(t - T)3
d. x(t) = (1/2)b(t - T)
e. x(t) = (1/3)b[(t - T)3 + T3]
f. None of the above

I was thinking d, but not sure.

doesn't look like it to me.

v(t) = dx(t)/dt

what is the derivative of option d?

ok when I did the derivative of vx(t)=b(t-T)^2, i got "a".

or do they want the antiderivative of that equation?

i'm not sure how to do the derivative of option d though because I'm not sure what to do about those constants. Wouldn't the derivative of b and T be 0 since they are constant?

courtney1121 said:
ok when I did the derivative of vx(t)=b(t-T)^2, i got "a".

or do they want the antiderivative of that equation?

i'm not sure how to do the derivative of option d though because I'm not sure what to do about those constants. Wouldn't the derivative of b and T be 0 since they are constant?

the derivative of a constant is zero. what the above equation i posted says is that the velocity (v) is the time derivative of position (x). if you differentiate the initial equation (velocity) you get the acceleration. that's not what you're after.

ooo ok ok...I need to get vx(t) in terms of x(t)

so i should take the antiderivative of the equation like use the reverse kinematic chain...so let me see here...

oooo ok ok I think I got this...I'm thinking the answer will be c...because if you take the derivative of c it would be (1/3)3b(t-T)^2...the 3's cancel out and you are left with b(t-T)^2 which is vx(t).

Does this sound right?

yes, looks correct to me.

## 1. What is the meaning of the variables in the velocity function?

The variable v represents the velocity, which is the rate of change of an object's position with respect to time. The variables t and T represent time, and b is a constant that affects the shape of the function.

## 2. How do you find the position function from the given velocity function?

To find the position function, you need to integrate the velocity function with respect to time. The resulting function will represent the position of the object at any given time t.

## 3. Can this velocity function be used for any type of motion?

Yes, this velocity function can be used for any type of motion as long as the acceleration is constant. This function is commonly used to model the motion of objects under the influence of gravity.

## 4. How does the value of the constant b affect the shape of the velocity function?

The constant b affects the shape of the velocity function by determining the rate at which the velocity changes with respect to time. A larger value of b will result in a steeper curve, while a smaller value of b will result in a gentler curve.

## 5. Can the given velocity function be used to calculate the position of an object at a specific time?

Yes, the position of an object at a specific time can be calculated by plugging in the value of t into the position function obtained by integrating the velocity function. This will give the exact position of the object at that particular time.

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