+2448 4. A rocket experiences an acceleration of 4 m/s^2 . How far has it traveled when it reaches 50 m/s?
I believe distance= 312.5 m
5. Your car has a maximum deceleration of 5 m/s^2 . What is your stopping distance if you were to slam on the brakes going 20 m/s?
Not sure how to solve this one >.<
I have to use this formula: vf^2=vi^2+2ad
vf=final velocity
vi=initial velocity
a=acceleration
d=distance
+36916 x=xo+ vot + 1/2 at^2 xo = 0 vo=0 a = 4
v=vo + at v=50 m/s vo=0 a = 4 m/s^2 solve for t = 12.5 s
x = 0 + 0 + 1/2 (4)(12.5)^2 = 312.5 m You got it! (Well.....at least we both arrived at the same answer!)
+36916 v=0 when car stops
v=vo + at vo= 20 m/s a=-5 m/s^2
0=20 +(-5)t
t= 4 seconds
x=xo + vot + 1/2 at^2
x = 0 + 20(4) + 1/2 (-5)(4)^2
= 80 + (-40)= 40 m (as Chris found)
+128475 Not great at Physics, but I'll give it a shot
vf^2 = vi^2 + 2ad
vi = initial velocity = 0....so we have
(50m/s)^2 = 0 + 2 (4m/s^2) * d
2500m^2 / s^2 = 8 m/s^2 * d divide both sides by 8m/s^2
[2500m^2/ s*2] / [ 8m /s^2] = d = 312.5 m ......correct !!!
Second one
vi = 20m/s
vf = 0 [ we are stopped at the end ]
a = -5m/s^2 [ we are slowing down ]
So
0 = [20m/s]^2 + 2(-5m/s^2) * d
0 = 400m^2 / s^2 - 10 m/s^2 * d multiply through by s^2
0 = 400m^2 - 10m * d
10m * d = 400m^2 divide both sides by 10m
d = 400m^2 / 10m = 40 m = stopping distance
