Kinematics, Displacement from Velocity-Time

[cidilon]

Homework Statement

Andrew is driving his van to deliver groceries in Elmvale. As he travels along Hwy 92, a fox stops on the road. Andrew is traveling at 75 km/h [North] and is 42 m away from the fox when he applies his brakes. It takes him 3.4 s to stop.

a) How far did Andrew travel before stopping?

v1 = -75 km/h [North] or -20.8 m/s [North] (It has to be negative because he is decelerating, right?)
v2 = 0
t = 3.4 s

Homework Equations

From looking at some examples, i believe i am supposed to use
d = .5(v2-v1)t + v1*t
or if i find acceleration first
d = v1*t + .5*a*t*t

The Attempt at a Solution

d = .5(v2-v1)t + v1*t
d = .5*(0+20.8)*3.4 - 20.8*3.4
d = -35.36 m [North] or -35 m [North]

As final velocity is known, can i use d = v2*t - .5(v2-v1)t or d = .5(v2+v1)t instead and does it matter?

What i am confused about is that d = .5(v2+v1)t is the area of a trapezoid, where as the slope as i imagine it for this word problem, makes a triangle. How is the area of a trapezoid working for a triangle? Does the slope NOT make a triangle? Should i not visualize this problem?

I really hope that i made myself clear, as it has been bugging me quite a lot!

 

[amy andrews]

All your calculations seem correct and so does your answer :)

v1 = -75 km/h [North] or -20.8 m/s [North] (It has to be negative because he is decelerating, right?)

This is not correct. The sign of acceleration does NOT affect the velocity. The sign of the velocity is only determined by its direction (since it's a vector). If you're going to give the velocity a sign, you'll have to determine which direction is positive (for example, west could be + and east, -) but in this case since there aren't different directions of velocities, that's pointless.

As final velocity is known, can i use d = v2*t - .5(v2-v1)t or d = .5(v2+v1)t instead and does it matter? !

If both equations are correct, and since acceleration is constant (some equations require constant acceleration), you should be able to use both to determine the distance.

What i am confused about is that d = .5(v2+v1)t is the area of a trapezoid, where as the slope as i imagine it for this word problem, makes a triangle. How is the area of a trapezoid working for a triangle? Does the slope NOT make a triangle? Should i not visualize this problem?

Distance will not be a slope, it will be an area on a graph just like work would be. It'll be velocity*time on a graph.
Hope this helps!

 

[cidilon]

All your calculations seem correct and so does your answer :)

This is not correct. The sign of acceleration does NOT affect the velocity. The sign of the velocity is only determined by its direction (since it's a vector). If you're going to give the velocity a sign, you'll have to determine which direction is positive (for example, west could be + and east, -) but in this case since there aren't different directions of velocities, that's pointless.

I was actually just thinking about this, it looked somewhat odd to me!
I got the idea in an example that i saw earlier. It was a similar problem but acceleration was found first. It went something like this "Kareem was driving his truck at 94 km/h [west]. He starts to apply brakes when he notice the stop sign ahead. It takes him 28 s to come to a complete stop"

For the answer on the other hand they used -94km/h [west] /28s = -0.93m/s2 [west] or 0.93m/s2 [east]. Why was the velocity made negative in this case? I understand that the acceleration needs to be negative because the object is decelerating. Is it okay to make it negative in case of acceleration?

Explain this to me and i think i will be good to go :) hehe

 

[amy andrews]

Yep, the acceleration will be negative since the velocity is going down (it's decelerating).
Glad to help ;)

 

[cidilon]

Wonderful, thanks a lot! :)