Question:

The time vs velocity diagrams of a 10-kg object for \(x\) and \(y\) directions are given as shown. Describe each motion for each time interval.

(1) 0 < \(t\) < 2 seconds

(2) 2 < \(t\) < 3 seconds

(3) 3 < \(t\) < 4 seconds

(4) at 2 seconds

Answer:

(1) From the diagram, the object moves at the constant velocity, 1 m/s, in \(x\) direction, but it gets acceleration of \(\frac{3}{2}\) m/s\(^2\) in \(y\) direction. The slope indicates the acceleration. Thus, it is exerted by 15 N in that direction because \(F=ma\).

(2) For both directions, during this time interval, the object moves at the constant velocity. The diagrams show the flat slopes; namely, no acceleration and no force on the object.

(3) The accelerations from 3 s to 4 s are

\[  a_x = a_y = \frac{0-3}{4-3} = -3 \ \mathrm{m/s^2}  \]

The magnitude of the negative acceleration in both directions is equal, so the force is directed in opposite to 45 degrees on \(x-y\) plane. The magnitude of force is given by

\[  |F| = 10 \times \sqrt{3^2+3^2} = 42 \ \mathrm{N}  \]

(4) At 2 seconds, the velocity in \(x\) direction is suddenly changed from 1 m/s to 3 m/s. This indicates the change in momenta; namely, there is an impulse in positive \(x\) direction. The impulse is calculated as

\[  I = mv_f – mv_i = 10 \cdot 3 – 10 \cdot 1 = 20 \ \mathrm{N\cdot s}  \]