Calculus-physics challenge!?

Question

These are some that have stumped me. Feel free to try any yo ucan. How do you find the position formula if you just have the velocity? Thanks

A projectilve is fired straight upward with a velocity of 400 ft/sec.
What is the position formula?
Find the time and the velocity when it hits the ground
Find the maximum altitude achieved by the projectile
Find the acceleration at any time

A point is moving on a coordinate line has position function s(t)=t/\3-9t+1 on [-3,3]
Find the velocity at time t
Find the acceleration at any time
When is it moving to the left?
when is it moving to the right?
When is it stopped?

A projectile is fired directly upward with position s(t)=100-192t-16t/\2
Find the velocity?
Find the acceleration?
Find the maximum height?
Find the duration of the flight?
When is the velocity zero?
What is the intial veloctiy?
What was the speed at impact?

Answer 1

1. Given a constant acceleration g then

dv/dt=g
v(t)=gt + V0 ; similarly
ds/dt=v=gt + V0
s(t)=0.5gt^2 + V0t + S0

2. Since s(t)=t^3-9t+1
v(t) ds/dt=3t^2 - 9
a=dv/dt= 6t
left or right? is when the slope
v=ds/dt>0 to the right
v=ds/dt<0 to the left
it will stop when v(t)=0 or 3t^2 - 9=0
t=sqrt(9/3)= sqrt(3)

3.
Find the velocity?
v=ds/dt
Find the acceleration?
a=dv/dt
Find the maximum height?
Find max
a) find first derivative and find the value of t for which ds/dt =0.
Find the duration of the flight?
t=? when V=0 and a=0
When is the velocity zero?
What is the initial velocity? see earlier solution
What was the speed at impact?

Answer 2

#1
The problem does say the projectile was fired straight up so I would assume that you know the velocity (given) and the acceleration (g). So the position formula would be
x(t) = x initial + 400 (t) + (1/2)(-9.81)(t)^2
The velocity when the projectile hits the ground will be the same as when it left the ground (400 ft/s) assuming no loss to air resistance.
To find the time it takes for the projectile to hit the ground. I would first differentiate both sides of the position function to get the velocity function.
v(t) = 400 + (-9.81)(t)
Now here you can do couple of different things to determine the time. 1) You could set the velocity function equal to zero and solve for t. This would give you the time it takes to reach its maximum height. You know that the acceleration of gravtiy is constant so its going to take the same amount of time to come back down so you multiply by two. or 2) You know that the projectile will have a velocity of -400 ft/s when it hits the ground so you just set the velocity function equal to -400 and solve for t.
The maximum height can be found by setting the velocity function equal to zero and solving for t. This is because the maximum height will be attained when the projectile has a velocity of zero. Then plug that number for t into the position function and solve for the position.
The acceleration at any time is g (-9.81 m/s^2)

#2
To find the velocity function you just differentiate both sides of the position function to get...
v(t) = 3t^2 -9
To find the acceleration you just differentiate both sides of the velocity function to get
a(t) = 6t
Using the velocity function you can see that the point is moving left when its velocity is negative and right when its positive.

moving left for t < sqrt 3 seconds
moving right for t > sqrt 3 seconds
stopped at t = sqrt 3

#3 can be solved using the same methods I have discussed in the previous examples

Hope this helps

Answer 3

1.
s=400t-1/2gt^2

hits the ground when s = 0

400t-1/2gt^2=0
t(800-gt)=0

t=0 (when it was thrown)
ot t=800/g secs (when it hits the ground)

ds/dt = v = 400-gt
at t=800/g

v=400-800 = -400 ft/s (positive velosity being upwards)

max altitude when v=0

400-gt=0
t=400/g secs
s=400x400/g - 1/2gx(400/g)^2
s=8000/g ft

acceleration = dv/dt=-g (upwards being positive)

you do the other 2

Answer 4

An electron is released from relax at a distance of 0.410m from a huge insulating sheet of value that has uniform floor value density 3.50×10?12C/m2 . A) How plenty artwork is achieved on the electron by making use of the electrical powered container of the sheet because of the fact the electron strikes from its preliminary place to a factor 7.00×10?2m from the sheet? B) what's the value of the electron while that's 7.00×10?2m from the sheet?