Question

A particle moves in two dimensions. The x-position in meters of a particle as a function of time in seconds is given by x(t) = 3 - 7t + 4t². The y-position in meters of the same particle as a function of time in seconds is given by y(t) = 1 + 2t +3t². If the particle's mass is 3.4kg, what force (magnitude and direction) is being applied?

Answer 1

x-position is given as,

x(t) = 3 - 7t + 4t^2

v(t) = dx(t)/dt = -7 + 4*2t

So, acceleration in the x-direction,

a(t) = dv(t)/dt = 4*2 = 8 m/s^2

Like-wise, acceleration in the y-direction is determined as -

y(t) = 1 + 2t + 3t^2

v(t) = dy(t)/dt = 2 + 3*2t

a(t) = dv(t)/dt = 3*2 = 6 m/s^2

So, the resultant acceleration of the particle is -

a = sqrt[8^2 + 6^2] = sqrt[100] = 10.0 m/s^2

Mass of the particle, m = 3.4 kg

Hence magnitude of the force, F = m*a = 3.4*10.0 = 34.0 N (Answer)

Direction of the force applied = direction of the acceleration.

=> = tan^-1(6/8) = 36.9 deg. from the positive direction of x - axis in counter-clockwise direction.