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 [tex]9.8 m/s^2[/tex] downward
f) See graphs in attachment
Explanation:
a)
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
[tex]a=g=-9.8 m/s^2[/tex]
downward (acceleration due to gravity). So, we can use the following suvat equation:
[tex]v=u+at[/tex]
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,
[tex]t=\frac{v-u}{a}=\frac{20-40}{-9.8}=2.04 s[/tex]
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
[tex]v=u+at[/tex]
And substituting
u = +40.0 m/s
a=g=-9.8 m/s^2
We find the corresponding time t:
[tex]t=\frac{v-u}{a}=\frac{-20-(+40)}{-9.8}=6.12 s[/tex]
c)
To solve this part, we can use the following suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s is the displacement
u = +40.0 m/s is the initial velocity
[tex]a=g=-9.8 m/s^2[/tex] 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,
[tex]0=ut+\frac{1}{2}at^2\\t(u+\frac{1}{2}a)=0[/tex]
which gives two solutions:
t = 0 (initial instant)
[tex]u+\frac{1}{2}at=0\\t=-\frac{2u}{a}=-\frac{2(40)}{-9.8}=8.16 s[/tex]
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
[tex]v=u+at[/tex]
where
u = +40.0 m/s is the initial velocity
[tex]a=g=-9.8 m/s^2[/tex] 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:
[tex]t=\frac{v-u}{a}=\frac{0-(40)}{-9.8}=4.08 s[/tex]
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
[tex]F=mg[/tex]
where m is the mass of the boulder and [tex]g[/tex] 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:
[tex]F=ma[/tex]
Combining the two equations, we get
[tex]ma=mg\\a=g[/tex]
So, the acceleration of the boulder is [tex]g=9.8 m/s^2[/tex] 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 [tex]-9.8 m/s^2[/tex], so this graph is a straight horizontal line.
Learn more about accelerated motion:
brainly.com/question/9527152
brainly.com/question/11181826
brainly.com/question/2506873
brainly.com/question/2562700
#LearnwithBrainly
