The position vector of a particle of mass 3 kg moving along a space curve is given by \(r = (4t^{3} - t^{2})i - (2t^{2} - t)j\) at any time t seconds. Find the force acting on it at t = 2 seconds.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
\(r = (4t^{3} - t^{2})i - (2t^{2} - t)j\)
\(v = \frac{\mathrm d r}{\mathrm d t} = (12t^{2} - 2t)i - (4t - 1)j\)
\(a = \frac{\mathrm d v}{\mathrm d t} = (24t - 2)i - 4j\)
\(t = 2 : a = (24(2) - 2)i - 4j\)
\(a = 46i - 4j\)
\(F = ma = 3(46i - 4j)\)
= \(138i - 12j)N\)
\(r = (4t^{3} - t^{2})i - (2t^{2} - t)j\)
\(v = \frac{\mathrm d r}{\mathrm d t} = (12t^{2} - 2t)i - (4t - 1)j\)
\(a = \frac{\mathrm d v}{\mathrm d t} = (24t - 2)i - 4j\)
\(t = 2 : a = (24(2) - 2)i - 4j\)
\(a = 46i - 4j\)
\(F = ma = 3(46i - 4j)\)
= \(138i - 12j)N\)