the graph shows the variation with time (t) of the force (f) acting on an object of mass 15000 kg. the…

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})

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})

Answer

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})