Calculating Position and Acceleration of a Test Car Under Microprocessor Control

[dovec]

The position of the front bumper of a test car under microprocessor control is given by x(t)=2.17m + (4.80m/s^2)t^2-(0.100m/s^6)t^6
Find its position at the first instant when the car has zero velocity.
Find its acceleration at the first instant when the car has zero velocity
Find its position at the second instant when the car has zero velocity
Find its acceleration at the second instant when the car has zero velocity
Please anyone, I need help with this, I'm not getting the correct answer! Thanks

 

[rock.freak667]

velocity=x'(t)
acceleration=v'(t)=x''(t)

you will need to use these identities.

 

[HallsofIvy]

The position of the front bumper of a test car under microprocessor control is given by x(t)=2.17m + (4.80m/s^2)t^2-(0.100m/s^6)t^6
Find its position at the first instant when the car has zero velocity.

Differentiate to find the velocity function. Set that equal to 0 and solve for t. There apparently will be more than one solution. Put the smallest positive solution into the x(t) function and evaluate.

Find its acceleration at the first instant when the car has zero velocity

Differentiate again to find the acceleration function. Put the t you found above into that acceleration function and evaluate.

Find its position at the second instant when the car has zero velocity

This is why I said there "apparently will be more than one solution". Set t equal to the next larger solution and evaluate x(t) for that t.

Find its acceleration at the second instant when the car has zero velocity
Please anyone, I need help with this, I'm not getting the correct answer! Thanks

Put that new value of t into the acceleration function and evaluate. If you have done these things and are not getting the correct answer, show what you have done.

 

[dovec]

x(t) = 0 + 2(4.80m/s^2)t - 6(0.100m/s^6)t^5
dx/dt = 2(4.80)t - 6(0.100)t^5=0
I keep getting t=16, but that's not right, please help me solve! please!

 

[tiny-tim]

Welcome to PF!

Hi dovec! Welcome to PF!

(try using the X² tag just above the Reply box )

2(4.80m/s^2)t - 6(0.100m/s^6)t^5
…
I keep getting t=16!

no you don't! put that equal to 0, and divide by t, and you get … ?

 

[HallsofIvy]

x(t) = 0 + 2(4.80m/s^2)t - 6(0.100m/s^6)t^5
dx/dt = 2(4.80)t - 6(0.100)t^5=0

This is not dx/dt, it's just x(t) again, with the initial "0" dropped!

I keep getting t=16, but that's not right, please help me solve! please!

If $x(t)= 9.6 t- .6 t^5$, then $dx/dt= 9.6- 3.0t^4$
Set that equal to 0 and you get $t^4= 9.6/3.0= 3.2$. Take the fourth root of 3.2. It is definitely NOT 16!

 

[dovec]

ok thanks a lot guys, I think I finally figured it out.