Question

2. A ball is kicked off a high hill. It’s horizontal and vertical positions as a function of time are given by x = 20.0t   and y = 2.00t − 4.90t2, where x and y are in meters and t is in seconds.

a) Write vector expressions for the ball's position (in m), the velocity in (m/s), the acceleration (in m/s2) as a function of time.

b) Write vector expressions for the position, the velocity, and the acceleration of the ball at t = 3.00 s.

please explain each step thanks

Answer 1

a)

along x axis

vx = dx/dt = 20 m/s

ax = dv/dt = 0 m/s^2

long y axis

vy = dy /dt = 2 - 9.8 t

ay = dvy/dt = - 9.8 m/s^2

now,

position vector

D = 20 t i + ( 2 t - 4.9 t^2) j

velocity vector

V = 20 i + ( 2 - 9.8 t) j

acceleration vector

a = - 9.8 j

==========

b)

position vector

D = 20 * 3 i + ( 2* 3 - 4.9* 3^2)j

D = 60 i - 38.1 j

velocity vector

v = 20 i - 27.4 j

acceleration vector

a = - 9.8j

=======

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