Question

Initially, a particle is moving at 5.60 m/s at an angle of 35.0° above the horizontal (+x axis). Two seconds later, its velocity is 5.24 m/s at an angle of 50.0° below the horizontal. What was the particle's average acceleration during these 2.00 seconds?
So I dont get the correct answers but I found the vector equations using trig function(sin,cos,tan).
For initial movement I got:
A=4.59m/s (x) +3.21 (Y)
After the two seconds:
B=4.01 (x) -3.36 (y)
then I thought I just subtract the x's and divide by two and do that same for y. Idk if im using the correct angle for the second motion

Answer 1

.

Answer 2

Well, your numbers don't quite match what I get for the second vector. What angle are you using for the second?

Answer 3

Oh, I see. I actually do get the "same" numbers, just not in the same order.

Your angle should be -50 degrees.
B = (5.24cos(-50), 5.24sin(-50)) = (3.37, -4.01)

Answer 4

Okay that makes sense. So I just add/subtract the X values of A and B and divide by two to get the acceleration for the x direction?

Answer 5

Yeah, you can find the acceleration in each direction, and then combine them in the normal way to find the total acceleration and angle.

Answer 6

so for the y vector acceleration I got Vf-Vi so -3.36-3.21 and divided by 2 to get -3.285 but it says its wrong

Answer 7

For the accelerations, I get:
\[a = (\frac{(3.37-3.37)}{2},\frac{-4.01-3.21}{2}) = (-0.61,-3.61) \]

Answer 8

I typed something wrong in the x acceleration

(3.37-4.59)/2

Answer 9

Oh my gosh thank you so much. I might just started mixing wrong number together. Thanks again!

Answer 10

My pleasure :)

That happens a lot. Just be careful! (Easier said than done, I know, haha!)