Question

suppose a tourist threw the rock with an initial velocity of 8.0m/s[up]. determine the velocity and displacement of the rock at 4.s​

Answer 1

1) The velocity is 31.2 m/s downward

2) The rock's displacement is 46.4 m downward

Explanation:

1)

The rock is in free fall (= acted upon gravity only), so its motion is a motion at constant acceleration, therefore we can use the following suvat equation:

$v=u+at$

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the rock in this problem, choosing upward as positive direction, we have:

u = +8.0 m/s

$a=g=-9.8 m/s^2$ (acceleration of gravity, downward)

So, the velocity at time

t = 4.0 s

is

$v=+8+(-9.8)(4)=-31.2 m/s$

And the negative sign means the direction is downward.

2)

To find the displacement we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement

u is the initial velocity

a is the acceleration

t is the time

In this situation:

u = +8.0 m/s

$a=g=-9.8 m/s^2$

t = 4.0 s

Substituting, we find

$s=(8.0)(4.0)+\frac{1}{2}(-9.8)(4.0)^2=-46.4 m$

And the negative sign means the displacement is downward.

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