How long for a cop car accelerating at 9 km/h² to catch a speeder going 110 km/h?

[anglum]

but what is that the time of? i am so lost right now

 

[learningphysics]

ok so when solving that i get

t = 24.81927711 seconds? is that correct?

yes.

 

[anglum]

what is that 24.8 the time of?\

and what steps do i take next i am so lost on this problem

 

[anglum]

o nevermind wow its late... that's the time it takes to catch up.. .i forgot what we were solving for jeeezzzz I am a moron

 

[learningphysics]

what is that 24.8 the time of?\

and what steps do i take next i am so lost on this problem

That's the time the one car catches up with the other... they start at the same time... on car quickly goes past the other with constant speed... the other car accelerates, and eventually goes faster than the other car... so it catches up..

Are there more parts?

 

[anglum]

ok so can i move onto the next problem i have...

a ball is thrown at an angle of 50degrees at 17m/s ,,,, how long does it take to reach its maximum height??

 

[learningphysics]

ok so can i move onto the next problem i have...

a ball is thrown at an angle of 50degrees at 17m/s ,,,, how long does it take to reach its maximum height??

What is its vertical velocity at its maximum height?

 

[anglum]

im not sure what that is... these are my toughest problems

 

[learningphysics]

im not sure what that is... these are my toughest problems

At the max. height, the vertical velocity is 0. What is the initial vertical velocity?

 

[anglum]

the initial velocity is 17m/s but its at a 50 degree angle so what would its vertical velocity be

 

[learningphysics]

the initial velocity is 17m/s but its at a 50 degree angle so what would its vertical velocity be

17sin(50)

 

[anglum]

ok so the vertical velocity is 13.022m/s? so then i can plug that into the

Vf= Vi - AT

and solve for T and that is the time to reach its vertical max?

 

[anglum]

this question has 2 more parts

this ball is shot by a person at a height of 2.626m and goes thru a hoop 3.048m high

what is the distance of the shot... and how long does it take to reach the hoop?

again its shot at 17m/s at angle of 50 degrees

 

[learningphysics]

What is the vertical displacement of the ball?

Use d = (vsin(theta))*t + (1/2)(-g)t^2 to find the time...

Then find the horizontal distance using vcos(theta)*t

 

[anglum]

what the heck is vsin? theta? -g?

the vertical displacement is .422 meters correct?

 

[learningphysics]

what the heck is vsin? theta? -g?

the vertical displacement is .422 meters correct?

Yes. Try to use the equation d = v1*t + (1/2)at^2 in the vertical direction... that's the equation I gave you v = 17. theta = 50. g = 9.8

 

[anglum]

but i don't know what d or t are so how do i solve for that?

 

[learningphysics]

but i don't know what d or t are so how do i solve for that?

d=0.422. solve for t.

 

[anglum]

but that won't be the answer for how long it takes the ball to travel the horizontal distance to the hoop... the hoop is x away from the person... how can i solve how long it takes to reach the hoop without knowing how far the hoop is from the person?

 

[learningphysics]

but that won't be the answer for how long it takes the ball to travel the horizontal distance to the hoop...

it is the same time.

the hoop is x away from the person... how can i solve how long it takes to reach the hoop without knowing how far the hoop is from the person?

It is the same time. 17cos(50)*t gives the horizontal distance.

 

[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)?