In what direction is it moving? A snowmobile is originally at the point with pos

Question

In what direction is it moving? A snowmobile is originally at the point with position vector 28.5 m at 95.0 degree counterclockwise from the x axis, moving with velocity 4.77 m/s at 40.0 degree. It moves with constant acceleration 1.71 m/s^2 at 200 degree. After 5.00 s have elapsed, find the following. In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.28 m. The mug slides off the counter and strikes the floor 1.80 m from the base of the counter. With what velocity did the mug leave the counter?

Explanation / Answer

3.solution

part a

When the acceleration vector "a" is constant, the velocity vector as a function of time is:

v(t) = v0 + at

In unit vector form, that's:

v(t) = (v0_x + (a_x)t)i + (v0_y + (a_y)t)j

It is given that v0 equals 4.77 m/s at 40.0°, so that means:

v0_x = (4.77 m/s)cos(40°) and v0_y = (4.77 m/s)sin(40°).

Also, it is given that "a" equals 1.71 m/s² at 200°, so that means:

a_x = (1.71 m/s²)cos(200°) and a_y = (1.71 m/s²)sin(200°).

So, plug in those values for v0_x, v0_y, a_x, a_y, and plug in "5.00 sec" for "t".

v(t) = (4.77m/s*cos40º + 1.71m/s²*cos200º*5s) i
+ (4.77m/s*sin40º + 1.71m/s²*sin200º*5s) j

v(t) = -4.38 m/s i + 0.141 m/s j

part 3b.

When the the acceleration vector "a" is constant, the position vector as a function of time is:

s(t) = s0 + (v0)t + ½at²

In unit vector form, that's:

s(t) = (s0_x + (v0_x)t + ½(a_x)t²)i + (s0_y + (v0_y)t + ½(a_y)t²)j

It is given that s0 equals 28.5 m at 95°, so that means: s0_x = (28.5 m)cos(95°) and s0_y = (28.5 m)sin(95°).

So plug in those values for s0_x and s0_y; and the values for v0_x, v0_y, a_x, a_y as in Part (a); and plug in "5.00 sec" for "t"

r = (28.5m*cos95º + 4.77m/s*cos40º*5s + ½*1.71m/s²*cos200º*(5s)²) i
+ (28.5m*sin95º + 4.77m/s*sin40º*5s + ½*1.71m/s²*sin200º*(5s)²) j
r = -4.29m i + 36.40m j

4a.solution

find the time for a 1.28 m free fall
s = 1/2 gt^2
t = (2s/g)
t = (2*1.28/9.81)
t = 0.510 s

find the final velocity for the free fall through 0.78 m
v = (2gs)
v = (2*9.81*1.28)
v = 5.01 m/s

find the horizontal velocity
the horizontal distance is 1.80 m in 0.510 s
v = s/t
v = 1,80 / 0.510 = 3.52 m/s = horizontal velocity leaving counter

part B

angle = tan^-1(vertical velocity/horizontal velocity)
angle = tan^-1(5.01 / 3.52)
angle = 54.90° <--- direction of final velocity below horizontal

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