Question

A force is applied to an 800 g object. Its velocity as a function of time is given v(t)= 2 [(t +1 s) m/s^2(i) − (t +1s)^-3 (ms^2)(j)].

(a) Find the magnitude of the initial velocity and its direction with respect to the positive x-direction.

(b) Find the magnitude of the force at the time of 1 s.

(c) Find the equation of the object’s trajectory. Initially, the object was at the origin.

Answer 1

v = 2 (t + 1)i - 2(t + 1)^-3 j

(A) at t = 0

v = 2i - 2j

magnitude = sqrt(2^2 + 2^2) = 2.83 m/s

direction = - tan^-1(2/2) = - 45 deg

(B) a = dv/dt = 2i + 6/(t+1)^4 j

at t = 1

a = 2i + 0.375 j

|a| = sqrt(2^2 + 0.375^2) = 2.03 m/s^2

F = m a = 1.63 N .....Ans

(b) r = integral of vdt

r = (t + 1)^2 i + (t + 1)^-2

x = (t + 1)^2

y = (t + 1)^-2

x y = 1 ......Ans