a) 2.75 s
The vertical position of the ball at time t is given by the equation
where
h = 4 m is the initial height of the ball
u = 12 m/s is the initial velocity of the ball (upward)
g = 9.8 m/s^2 is the acceleration of gravity (downward)
We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:
This is a second-order equation. By solving it for t, we find:
t = -0.30 s
t = 2.75 s
The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.
b) -15.0 m/s (downward)
The final velocity of the ball can be calculated by using the equation:
where
u = 12 m/s is the initial (upward) velocity
g = 9.8 m/s^2 is the acceleration of gravity (downward)
t is the time
By subsisuting t = 2.75 s, we find the velocity of the ball as it reaches the ground:
And the negative sign means the direction is downward.
Answer:
120 N
Explanation:
F=ma therefore 60kg times 2m/s^2 is 120 N
Answer: d. 5 m/s^2
Explanation:
Acceleration is the change in velocity in a given time.
a = (30-20)/2 = 5
Answer:
40000÷40=1000 joules is required to work in 40 seconds
<span>You can use the equation
V_xf = V_xi + a_x(t)
V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0
t
Thus, solve for t and get 10seconds
and then take 5 seconds to break after 20 seconds of driving
so for
a) 10 + 20 + 5 = 35 seconds
</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity
Need to find xf, knowing xi = 0
Thus, use the formula
x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
x_f = 100m
so for the first 10 seconds the truck traveled 100ms
At a speed of 20m/s
20m/s = xm/20s
20*20 = x
x = 400
thus we have 100+400 = 500m
then it slows down from 500m to x_f
thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5)
x_f = 500 + 50
x_f = 550
therefore the total distance traveled is 550m
</span>
<span>to calculate average velocity
550/35 = 16m/s
thus
V_xavg = 16m/s</span>
