Question

the graph shows the variation with time (t) of the force (f) acting on an object of mass 15000 kg. the object is at rest at (t = 0). what is the speed of the object when (t=30 s)? a (0.18 m s^{-1}) b (6 m s^{-1}) c (12 m s^{-1}) d (180 m s^{-1})

Explanation:

Step1: Recall impulse - momentum theorem

Impulse \(J=\Delta p\), where \(J=\int Fdt\) and \(\Delta p = m\Delta v\). For a non - zero initial momentum \(p_0 = 0\) (object at rest initially), \(J = mv\). The impulse is equal to the area under the \(F - t\) graph.

Step2: Calculate the area under the \(F - t\) graph

The \(F - t\) graph is a triangle. The area of a triangle \(A=\frac{1}{2}\times base\times height\). Here, the base of the triangle is \(t = 30s\) and the height is \(F=6000N\) (since \(F\) is in kN, \(6kN = 6000N\)). So, \(J=\frac{1}{2}\times30s\times6000N=90000Ns\).

Step3: Use the impulse - momentum relation to find velocity

We know \(J = mv\), and \(m = 15000kg\). Rearranging for \(v\), we get \(v=\frac{J}{m}\). Substituting \(J = 90000Ns\) and \(m = 15000kg\) into the formula, \(v=\frac{90000Ns}{15000kg}=6m/s\).

Answer:

B. \(6m s^{-1}\)