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Laws of motion problem

  • Thread starter Leesh09
  • Start date
  • #1
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Homework Statement



To model a spacecraft, a toy rocket engine is securely fastened to a 2.00-kg large puck, which can glide with negligible friction over a horizontal surface, taken as the xy plane. The engine exerts a time-dependent force, F = (8.00 î – 4.00t ĵ) N, where t is in seconds, on the puck. If the puck is initially at rest, (a) At what time will it be moving with a speed of 15.0 m/s? (b) How far is the puck from its initial position when its speed is 15.0 m/s? (c) Through what total displacement has the puck traveled at this time?

Homework Equations



Fx=ma, Fy=ma, v=at

The Attempt at a Solution



we know m=2.00 kg and F=(8.00i-4.00tj)
For the x component of F, 8.00 N= (2.00 kg)(x component of acceleration)
so a sub x= 4.00 m/s^2.
For the y component: -4.00t=(2.00)(y component of a) so a sub y= -2.00t m/s^2

To find the time at which the speed is 15.0 m/s I would think you would take v=(ax+ay)t and then set v=to 15 so 15=(4.00+(-2.00t))*t so 15=4.00t-2.00t^2. This is a dead end, however, since at no point does the equation -2.00t^2+4.00t-15=0/
 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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Welcome to PF!

Hi Leesh09! Welcome to PF! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)
To find the time at which the speed is 15.0 m/s I would think you would take v=(ax+ay)t …
No, vx= axt, and vy= ayt,

but v is not vx + vy

how do we combine velocities in perpendicular directions? :smile:
 
  • #3
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0
would it work to say a(t)=(4.00i-2.00tj), take the integral to find velocity and then find this magnitude? so 15= sqrt(4.00t2+(-t2)2)?? so then 225=16t2+t4, getting t=3 s?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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… so 15= sqrt(4.00t2+(-t2)2)??
erm :redface: … where did the (t2)2 come from?
 

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