Solve Car Chase Question: 110 km/h and 170 km/h

[anglum]

ok so if i solve for t on that last equation u gave me...

.422= 17sin50 (t) = .5(-9.8)t squared

.422 = 13.022t + -4.9t squared

and then i get stuck in solving for t

 

[anglum]

as u can tell my algebra is strugglin tonite

 

[learningphysics]

ok so if i solve for t on that last equation u gave me...

.422= 17sin50 (t) = .5(-9.8)t squared

.422 = 13.022t + -4.9t squared

and then i get stuck in solving for t

use the quadratic equation.

 

[anglum]

god i feel so dumb right now ... how do i use the quadratic equation to solve that

this is bad ... and I am so sorry

 

[learningphysics]

as u can tell my algebra is strugglin tonite

no prob.

 

[anglum]

i can't even function as to the quadratic equation solving that for me...

 

[learningphysics]

god i feel so dumb right now ... how do i use the quadratic equation to solve that

this is bad ... and I am so sorry

that's ok. have a look at the quadratic formula here:

http://en.wikipedia.org/wiki/Quadratic_formula#Quadratic_formula

First arrange the equation in the correct form:

4.9t^2 -13.022t +0.422 = 0

now try to apply the quadratic formula using the formula in the link.

 

[anglum]

t = 2.62?

 

[learningphysics]

t = 2.62?

yes. looks right.

 

[anglum]

so then for the horizontal distance it is just 17 cos 50 (2.62)?

 

[learningphysics]

so then for the horizontal distance it is just 17 cos 50 (2.62)?

yes.

 

[anglum]

ok 2 more problems if u don't mind?

#1

a rotating cylinder 10 miles long and 4.9 miles in radius is in space...

acceleration of gravity is 9.8m/s squared

what angular speed must the cylinder have so that the centripetal acceleration at its surface equals the free fall acceleration on earth?

 

[anglum]

ok i got that problem on my own actually WOOO HOOO

now the 2nd problem if u don't mind

 

[anglum]

014 (part 1 of 2) 10 points
The speed of a point on a rotating turntable,
which is 0.131 m from the center, changes at a
constant rate from rest to 0.958 m/s in 1.95 s.
At t1 = 1.6 s, find the magnitude of the
tangential acceleration. Answer in units of
m/s2.
015 (part 2 of 2) 10 points
At t1 = 1.6 s, ¯find the magnitude of the total
acceleration of the point. Answer in units of
m/s2.

 

[anglum]

learning these are my last 2 problems ... sorry to keep bothering you

 

[learningphysics]

014 (part 1 of 2) 10 points
The speed of a point on a rotating turntable,
which is 0.131 m from the center, changes at a
constant rate from rest to 0.958 m/s in 1.95 s.
At t1 = 1.6 s, find the magnitude of the
tangential acceleration. Answer in units of
m/s2.

This is just the change in velocity divided by time. ie 0.958/1.95.

015 (part 2 of 2) 10 points
At t1 = 1.6 s, ¯find the magnitude of the total
acceleration of the point. Answer in units of
m/s2.

What is the centripetal acceleration?

 

[learningphysics]

ok 2 more problems if u don't mind?

#1

a rotating cylinder 10 miles long and 4.9 miles in radius is in space...

acceleration of gravity is 9.8m/s squared

what angular speed must the cylinder have so that the centripetal acceleration at its surface equals the free fall acceleration on earth?

use centripetal acceleration = v^2/r. At a radius r, v = rw.