How Far Will a Cart Travel Up a Ramp at Different Angles?

[Tonik]

Homework Statement

A cart is pushed up a ramp with an initial speed v₀=2.5m/s. For this problem, you may assume that the cart is frictionless and the acceleration of the cart a=gsin$$\theta$$.

A) If the angle of the ramp is $$\theta$$=20, what is the maximum distance d that the cart will travel up the ramp?

B) If the angle of the ramp is $$\theta$$=45, what is the maximum distance d that the cart will travel up the ramp?

Homework Equations

v²=v₀²+2a$$\Delta$$X
manipulated to...
(v²-v₀²) / (2a) = $$\Delta$$X

The Attempt at a Solution

A)
9.8sin(20)=3.351797
(0-(2.5)²) / (2(3.351797) = $$\Delta$$X
$$\Delta$$X=10.474 meters

B)
9.8sin(45)=6.9296
(0-(2.5)²) / (2(6.9296) = $$\Delta$$X
$$\Delta$$X=.4509 meters

I was feelin' good about the first answer but the second answer seems too low... did I do these correctly?

 

[Tonik]

Shameless bump.

 

[PhanthomJay]

Shameless bump.

B looks good. Just check your math error in A.

 

[Tonik]

B looks good. Just check your math error in A.

A)
9.8sin(20)=3.351797
(0-(2.5)²) / (2(3.351797) = $$\Delta$$X
$$\Delta$$X=.93234 meters

Fixed, I have no idea how I came up with ~10 meters...
Thanks bud!

 

[rwisz]

Good job on using the appropriate equations and solving for the maximum distance traveled by the cart. Your calculations and answers for part A seem correct. However, for part B, your answer of 0.4509 meters seems too low. When solving for the maximum distance, it is important to consider the initial velocity of the cart, which in this case is 2.5 m/s. This means that the cart has some initial velocity that will contribute to its maximum distance traveled. Therefore, your answer for part B should be larger than your answer for part A. I suggest double-checking your calculations and making sure you are using the correct values for the initial velocity and the acceleration. Keep up the good work!