
#1
Sep3009, 12:46 AM

P: 1

(1) 1. The problem statement, all variables and given/known data
Determine the stopping distance for a car with an initial speed of 95km/h and human reaction time of 1.0s, for an acceleration (a) a = 4.0m/s^2 Initial velocity (V1) = 95 km/h = 26.38 m/s Final velocity (V2) = 0 m/s Initial Displacement (Y1) = ? Final Displacement (Y2) = 0 m Acceleration = 4.0 m/s^2 2. Relevant equations V2 = V1 + at (??) Y2 = Y1 + V1(t) + 1/2(a)(t)^2 3. The attempt at a solution Tried too many times...  (2) 1. The problem statement, all variables and given/known data A helicopter is ascending vertically with a speed of 5.20 m/s. At a height of 125 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [ Hint: the package's initial speed equals the helicopter's.] 2. Relevant equations ??? 



#2
Sep3009, 12:52 AM

P: 5

First count the distance that it travels in one second (before the brake is applied). Then using the given acceleration find the distance it travels while decelerating from 95 to 0, you should use the equation: final v squared minus initial v squared is 2ad where a is acceleration and d is distance. Try that out.




#3
Sep3009, 01:02 AM

P: 42

So you have initival velocity 26.833m/s
Takes a human 1 second to hit the break; so find the distance traveled before he hits the break Formula:[tex]d = d_{o} + v_{0}*t+\frac{1}{2}at^2[/tex] Variables: [tex]d_{0} = 0[/tex] a = 0 [tex] v_{0} [/tex] = 26.83m/s t=1 Solve for d  Next find the time it takes for the driver to come to a stop Formula : [tex]V= V_{0} + at[/tex] Variables: V = 0 [tex]V_{0}[/tex] = 26.83m/s a = [tex]4m/s^2[/tex] solve for t  Next find the distance when he hits the break Formula;[tex]d = d_{o} + v_{0}*t+\frac{1}{2}at^2[/tex] Variables: [tex]d_{0}[/tex] = the answer from before [tex]v_{0}[/tex] = you already have a = [tex]4m/s^2[/tex] t = from before Solve for d 



#4
Sep3009, 01:10 AM

P: 39

A Couple Deceleration Questions..v^2  u^2 = 2as v = 0 (the cars stopped) u = 26.38m/s a = 4m/s^2 s is the unknown: this gives s = 86.98m now the driver takes 1s to put the brakes on, so distance travelled in one sec = u*t=26.38m. So total s = 86.98+26.38 = 113.36m For the second part, the package will first slow down to zero velocity and then accelerate downwards. So its like this: v=u+at; v^2u^2=2as; here v = 0 (final velocity where it stops, at the maximum height that it attains) u = 5.20m/s upwards a = 9.8m/s^2 downwards this gives t = 0.5306s, and s = 1.379m Now on its downwards motion v^2u^2=2as; here u = 0, a = 9.8m/s^2 downwards, s = 125+1.379=126.379m v^2=2477.0284 v = 49.769m/s (this is the impact velocity with which the package hits the ground) Time taken for this is v = u+at u = 0, v = 49.769, a = 9.8m/s^2 this gives t = 5.078s So the total time for the package is t = 5.078+0.5306=5.6086s here's an overview of how i have calculated this: the package first moves up with the helicopters speed till it reaches a maximum height and a zero velocity. From here, it falls freely under gravity. So i've computed the time for it to rise to a max height and the time it takes to fall from that position to the ground, add them up and thats the answer. 


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