Solving a 4 kg Particle's Velocity & Force

[tjbateh]

Homework Statement

The velocity of a 4 kg particle is given by v = (3ti + 6t^2j ) m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 38 N, what are the direction (relative to the positive direction of the x axis) of (a) the net force and (b) the particle's direction of travel?

Homework Equations

The Attempt at a Solution

I took the second derivative of V to get the acceleration which is 3(i) + 12t(j). What should my next step be?

 

[rock.freak667]

Force=ma

so get that vector and use the fact that the magnitude is 38N to find the time.

 

[tjbateh]

so would that vector be (12(i) + 12t(j))? And I would set that equal to 38?

 

[rock.freak667]

so would that vector be (12(i) + 12t(j))? And I would set that equal to 38?

no you'd get F=12i+48tj

and |12i+48tj|=38. Find t

 

[tjbateh]

how did you get 48t??

 

[rock.freak667]

how did you get 48t??

a=3i + 12tj

m=4, finding ma, 12*4=48. Unless I am wrong as I am bad with numbers.

 

[tjbateh]

ok, so T = .54 seconds. what should I do next?

 

[rock.freak667]

ok, so T = .54 seconds. what should I do next?

find the force vector (by subbing t=0.54)

then find the angle it makes with the x-axis. It's a similar process for part b

 

[tjbateh]

ok, the force vector is F= 1.62i + 1.75j
And the angle is 47.21 degrees? 47.21 is wrong though, i tried it.