Posted by **Segun Akowonjo** on Friday, February 14, 2014 at 8:22pm.

A snowmobile is originally at the point with position vector 28.7 m at 95.0° counterclockwise from the x axis, moving with velocity 4.85 m/s at 40.0°. It moves with constant acceleration 1.94 m/s2 at 200°. After 5.00 s have elapsed, find the following:

(a)Its velocity vector

(b)Its position vector

- physics -
**Steve**, Friday, February 14, 2014 at 11:14pm
s0 = 28.7 @ 95.0° = -2.501,28.591

v0 = 4.85 @ 40.0° = 3.715,3.119

a = 1.94 @ 200° = -1.823,-0.664

Now, we just plug in our usual equations:

v = v0 + at

s = s0 + v0*t + 1/2 at^2

and then convert back to polar form as needed.

## Answer This Question

## Related Questions

- physics - A snowmobile is originally at the point with position vector 26.1 m at...
- physics - A snowmobile is originally at the point with position vector 31.3 m at...
- physics - A snowmobile is originally at the point with position vector 31.5 m at...
- physics - A snowmobile is originally at the point with position vector 31.5 m at...
- Physics - A particle moves in the xy plane with constant acceleration. At time ...
- Physics - A particle moves along the x axis. It is initially at the position 0....
- physics - A particle moves along the x axis. It is initially at the position 0....
- physics - A small object with mass 4.05 kg moves counterclockwise with constant ...
- Physics - The acceleration of a particle moving only on a horizontal xy plane is...
- Physics 141 - A snowmobile moves according to the velocity-time graph shown in ...

More Related Questions