Find the rider's position and velocity after 1/4s.?

Question

a motorcycle stunt rider rides off the edge of a cliff with a horizontal velocity of magnitude 5m/s.

Answer 1

Use the positon equations:

Sx = Sox + Vox t + (1/2) ax t^2

Sy = Soy + Voy t + (1/2) ay t^2

For his horizontal position (Sx) , there is no acceleration so ax =0.  We can assume that the endge of the cliff is our reference point so Sox (initial positon in x) is 0 as well.

This gives us: Sx = 0 + Vox t + 0 = 5m/s * 0.25s = 1.25 m

For your vertical position (Sy), we can say that Soy = 0, we know that Voy = 0 (the bike wasn't moving down before he left the cliff) and ay = g = -9.81 m/s^2 (gravitational accelearation)

This gives us Sy = 0 + 0 + (1/2) (9.81m/s^2) (0.25s)^2 = -0.31 m

His position is (1.25m, -0.31m)

Answer 2

x-direction:

Vi = 5 m/s

Vf = ?

x = ?

a = 0 m/s²

t = 0.25s

y-direction:

Vi = 0 m/s

Vf = ?

y = ?

a = 9.8 m/s²

t = 0.25s

equations we will use:

x = Vi * t + (1/2) * a * t²

Vf = Vi + a* t

we can find the horizontal position by using the first equation with the x-direction information:

x = (5)(0.25) + (1/2)(0)(0.25)²

x = (5)(0.25) + 0

x = 1.25m

to find the vertical position, use the first equation with the y-direction information:

y = (0)(0.25) + (1/2)(9.8)(0.25)²

y = 0 + (1/2)(9.8)(0.0625)

y = 0.306m

actual position = √(x² + y²)

√[(1.25)² + (0.306)²] = 1.287m <---distance

tan(θ) = y / x

θ = tanˉ¹(y / x)

θ = tanˉ¹(0.306 / 1.25) = 13.76° south of east <---angle

to find the riders velocity we will use the second equation with the y-information:

Vf = Vi + a * t

Vf = (0) + (9.8)(0.25) = 2.45 m/s

and now with the x information:

Vf = Vi + a * t

Vf = (5) + (0)(0.25) = 5m/s

actual velocity = √(x² + y²)

√(2.45² + 5²) = 5.568m/s <---velocity answer

TO SUM IT UP:

position: 1.287m, 13.76° south of east

velocity: 5.568 m/s

hope that helped!