A 30g mathematically inclined mouse runs across teh flooraccording to the follow

Question

A 30g mathematically inclined mouse runs across teh flooraccording to the following equation. In this equation, trepresents the time in seconds from the start and s(t) representsthe displacement in meters as a function oftime.                           s(t)=[(9t)i+ (-2t^2)j] meters A) What is the magnitude and direction of the mouse's netdisplacement at the end of 2 seconds? B) What is the magnitude and direction of the mouse'sinstantaneous velocity at 2 seconds? C) What is the magnitude and direction of the mouse'sacceleration at t=2 seconds? D) What is the magnitude of teh force on the mouse at t=2seconds? Is the force constant? How do you know? A 30g mathematically inclined mouse runs across teh flooraccording to the following equation. In this equation, trepresents the time in seconds from the start and s(t) representsthe displacement in meters as a function oftime.                           s(t)=[(9t)i+ (-2t^2)j] meters A) What is the magnitude and direction of the mouse's netdisplacement at the end of 2 seconds? B) What is the magnitude and direction of the mouse'sinstantaneous velocity at 2 seconds? C) What is the magnitude and direction of the mouse'sacceleration at t=2 seconds? D) What is the magnitude of teh force on the mouse at t=2seconds? Is the force constant? How do you know?

Explanation / Answer

displacement S ( t ) = 9t i -2t^2 j (A). att= 2 s , S ( t ) = ( 9 * 2 ) i - ( 2 * 2 ^ 2 ) j                                = 18 i - 8 j magnitude S = [ 18 ^ 2 + -8 ^ 2 ]                    = 19.69 m Let the displacement makes an angle with positive x -axis then   tan = -8 / 18                                                                                                           = -23.96 degrees (B). velocity v = dS(t) / dt                       = 9 i - 4 t j Velocity at time t= 2 s is   v = 9i - 8j magnitude v = [ 9 ^ 2 + -8 ^ 2 ]                    = 12.041 m / s Let the velocity v makes an angle with positive x- axisthen tan = -8 / 9                                                                                                   = -41.63 degrees (c). Accleration a = dv / dt                            = d/ dt [ 9 i -4t j ]                            = -4 j magnitude = 4 m / s ^ 2 direction is in negative y direction (d). magnitude of force F = ma = 30 g * 4 m / s ^ 2=0.03 kg * 4 m / s ^ 2= 0.12 N direction is in negative y direction

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