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avatar+2442 

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

 Sep 13, 2018
 #1
avatar+17331 
+3

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!)

 Sep 13, 2018
edited by Guest  Sep 13, 2018
 #2
avatar+2442 
+1

Thanks :)

RainbowPanda  Sep 13, 2018
 #5
avatar+17331 
+2

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) 

ElectricPavlov  Sep 13, 2018
 #3
avatar+98125 
+2

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

 

 

cool cool cool

 Sep 13, 2018
 #4
avatar+2442 
+1

Thank you so much!

RainbowPanda  Sep 13, 2018

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