the last one Thanks A snowmobile is originally at the point with position vector

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the last one

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A snowmobile is originally at the point with position vector 31.7 m at 95.0 degree counterclockwise from the x axis, moving with velocity 4.65 m/s at 40.0 degree . It moves with constant acceleration 2.10 m/s2 at 200 degree . After 5.00 s have elapsed, find the following. its velocity vector its position vector An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 16.0 m if her initial speed is 2.20 m/s. What is the free-fall acceleration on the planet? (Ignore air resistance.) In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is h. The mug slides off the counter and strikes the floor at distance d from the base of the counter. With what velocity did the mug leave the counter? (Use any variable or symbol stated above along with the following as necessary: g.) Vxi = What was the direction of the mug's velocity just before it hit the floor? (Use any variable or symbol stated above as necessary.) 6 = below the horizontal A particle initially located at the origin has an acceleration of a = l.oof m/s2 and an initial velocity of vj = 8.001 m/s. Find the vector position of the particle at any time t (where t is measured in seconds). Find the velocity of the particle at any time t. Find the coordinates of the particle at t Find the speed of the particle at t = 4.00 s.

Explanation / Answer

1.

represent all of them in X nad y component. you will get

S = -2.76 i + 31.58 j

v = 3.562 i + 2.9889 j

a = -1.973 i - 0.718 j

now apply usual formula for velosity and displacement.

weknow,

V = u + ft

= -6.303 i - 0.6011 j m/S

S = ut + 0.5at^2

= -6.8525 i - 5.9861 j m.

so nw position vector r = addition of the initial position vector + the dispacement = - 9.6125 i - 25.594 j m.

you canvert them to polar form using cos() + j sin() formula.

2. see at the highest height his velocity will be zero. soif we use the following formula

V^2 = U^2 - 2*f*S where V is the final velocity, u the iitial velocty f is the free fall acceleration.

then putting V = 0 we get

f = 0.15125 m/S^2.

3.

I could not understand the problem

4.

just use the following formula

V = U+f*t

S = u*t + 0.5*f*t^2

use all the vector notation as it is and find the required values in vctor form.

in the last two part initially it has 0 posotion vector so what ever comes from the second equation that would be result. for velocity just add the caculated velocity with the initial velocity vector wise.

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