Solving a Cart Motion Problem: Understanding the Graphs

[joemama69]

Homework Statement

A .5 kg cart moves on a horizontal surface. The graph of v_x,t is attached

a) Indicate all time cart is at rest
b)indicate every time interval for which the speed is increasing
c)determine the horizontal position of x of the car at t=9s if the cart is located at x=2 @ t=0
d)sketch the acceleration versus time graph for the motion of the cart from t=0 to t = 2.5s
e)from t=2.5 until the car reaches the end of the trach, the car continues wiht constant horizontal velocity. the car leaves the end of the trach and htis the floor, which is .4m below the track. 1) find the time it takes to hit the floor from the track 2) find horizontal distance 3) the velocity when it hits the ground

Homework Equations

The Attempt at a Solution

a) t = 4 & 18
b) t= (9-12) & (17-20)
c)first i found the slop of the velocity m = (.8--1)/(0-9) = -1/5

so v = -1/5*t + .8

to find the displacement i believe i integrate to get

x = -1/10 t² + .8t

x(9) = -.9 where did i go wrong

d) it would simply be a horizontal line at a = -.2 from 0-9, then a horizontal line @ a = .2 from 9-12, then a horizontal ilne at zero from 12 - 17, then a horizontal line @ a = .4 from 17 to 20, then a horinzontal straight ine @ a = 0 from 20 +

e1) v_x = .8
y = y_o + v_yt - .5gt²
-.4 = -.5(9.8)t² t = 2.86 s

e2) x = x_o + v_xt
x = .8(2.86) = .23 m

e3) v_y = v_yo - gt
v_y = -9.8(2.86) = -28 m/s

so the initial velocity = sqrt(28² + .8²) = 28.01 m/s

 

[tiny-tim]

Hi joemama69!

… The graph of v_x,t is attached …

erm … nooo, it isn't!

 

[joemama69]

now it is sorry

 

[tiny-tim]

Hi joemama69!

ok, I can see it now!

Your a b and d are fine.

In your c, you've done two things wrong …

x = -1/10 t2 + .8t

x(9) = -.9 where did i go wrong

i] you didn't include the "constant" (the cart is located at x=2 @ t=0)

ii] you're completely missing the point of the graphical method by resorting to equations …

the advantage of a velocity/time graph is that you can find the distance by simply measuring the area under the graph (remember, of course, that "under" means "between the graph and the axis", and so is negative if the graph is below the axis )

(and you should have used the same method in e)