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

[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.

 

[teclo]

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?

 

[courtney1121]

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?

 

[teclo]

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.

 

[courtney1121]

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?

 

[teclo]

yes, looks correct to me.