Question

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. (a) At what time after being ejected is the boulder moving at 20.0 m/s upward? (b) At what time is it moving at 20.0 m/s downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch ay-t, vy-t, and y-t graphs for the motion.

Answer 1

a) Time at which velocity is +20.0 m/s: 2.04 s

b) Time at which velocity is -20.0 m/s: 6.12 s

c) Time at which the displacement is zero: t = 0 and t = 8.16 s

d) Time at which the velocity is zero: t = 4.08 s

e) i) ii) iii) The acceleration of the boulder is always $9.8 m/s^2$ downward

f) See graphs in attachment

Explanation:

a)

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

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

downward (acceleration due to gravity). So, we can use the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u = 40.0 m/s is the initial velocity

a=g=-9.8 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 20.0 m/s

Therefore,

$t=\frac{v-u}{a}=\frac{20-40}{-9.8}=2.04 s$

b)

In this case, we want to find the time t at which the boulder is moving at 20.0 m/s downward, so when

v = -20.0 m/s

(the negative sign means downward)

We use again the suvat equation

$v=u+at$

And substituting

u = +40.0 m/s

a=g=-9.8 m/s^2

We find the corresponding time t:

$t=\frac{v-u}{a}=\frac{-20-(+40)}{-9.8}=6.12 s$

c)

To solve this part, we can use the following suvat equation:

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

where

s is the displacement

u = +40.0 m/s is the initial velocity

$a=g=-9.8 m/s^2$ is the acceleration

t is the time

We want to find the time t at which the displacement is zero, so when

s = 0

SUbstituting into the equation and solving for t,

$0=ut+\frac{1}{2}at^2\\t(u+\frac{1}{2}a)=0$

which gives two solutions:

t = 0 (initial instant)

$u+\frac{1}{2}at=0\\t=-\frac{2u}{a}=-\frac{2(40)}{-9.8}=8.16 s$

which is the instant at which the boulder passes again through the initial position, but moving downward.

d)

To solve this part, we can use again the suvat equation

$v=u+at$

where

u = +40.0 m/s is the initial velocity

$a=g=-9.8 m/s^2$ is the acceleration

We want to find the time t at which the velocity is zero, so when

v = 0

Substituting and solving for t, we find:

$t=\frac{v-u}{a}=\frac{0-(40)}{-9.8}=4.08 s$

e)

In order to evaluate the acceleration of the boulder, let's consider the forces acting on it.

If we neglect air resistance, there is only one force acting on the boulder: the force of gravity, acting downward, with magnitude

$F=mg$

where m is the mass of the boulder and $g$ the acceleration of gravity.

According to Newton's second law, the net force on the boulder is equal to the product between its mass and its acceleration:

$F=ma$

Combining the two equations, we get

$ma=mg\\a=g$

So, the acceleration of the boulder is $g=9.8 m/s^2$ downward at any point of the motion, no matter where the boulder is (because the force of gravity is constant during the motion).

f)

Find the three graphs in attachment:

- Position-time graph: the position of the boulder initially increases as the boulder goes upward; however, the slope of the curve decreases as the boulder goes higher (because the velocity decreases). The boulder reaches its maximum height at t = 4.08 s (when velocity is zero), then it starts going downward, until reaching its initial position at t = 8.16 s

- Velocity-time graph: the initial velocity is +40 m/s; then it decreases linearly (because the acceleration is constant), and becomes zero when t = 4.08 s. Then the velocity becomes negative (because the boulder is now moving downward) and its magnitude increases.

- Acceleration-time graph: the acceleration is constant and it is $-9.8 m/s^2$, so this graph is a straight horizontal line.

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