How to Solve a Time-Dependent Force Problem for a Gliding Puck

[Leesh09]

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/

 

[tiny-tim]

Welcome to PF!

Hi Leesh09! Welcome to PF!

(try using the X² and X₂ tags just above the Reply box )

To find the time at which the speed is 15.0 m/s I would think you would take v=(ax+ay)t …

No, v_x= a_xt, and v_y= a_yt,

but v is not v_x + v_y …

how do we combine velocities in perpendicular directions?

 

[Leesh09]

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?

 

[tiny-tim]

… so 15= sqrt(4.00t²+(-t²)²)??

erm … where did the (t²)² come from?