(a) Forces \(F_{1} = \begin{pmatrix} -5 & 4 \end{pmatrix} N; F_{2} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} N; F_{3} = \begin{pmatrix} 2 & -1 \end{pmatrix} N\) and \(F_{4} = \begin{pmatrix} 3 & -5 \end{pmatrix} N\) act on a body. Find the :
(i) resultant of these forces ; (ii) fifth force that will keep the body in equilibrium.
(b) A body moving at 20 ms\(^{-1}\) accelerates uniformly at 2.5 ms\(^{-2}\) for 4 seconds. It continues the journey at the speed for 8 seconds, before coming to rest in t seconds with a uniform retardation. If the ratio of the acceleration to the retardation is 3 : 4,
(i) sketch the velocity- time graph of the journey ; (ii) find t ; (iii) find the total distance of the journey.
(i) resultant of these forces ; (ii) fifth force that will keep the body in equilibrium.
(b) A body moving at 20 ms\(^{-1}\) accelerates uniformly at 2.5 ms\(^{-2}\) for 4 seconds. It continues the journey at the speed for 8 seconds, before coming to rest in t seconds with a uniform retardation. If the ratio of the acceleration to the retardation is 3 : 4,
(i) sketch the velocity- time graph of the journey ; (ii) find t ; (iii) find the total distance of the journey.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option
Explanation:
(a) (i) \(F_{1} = \begin{pmatrix} -5 & 4 \end{pmatrix} N; F_{2} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} N; F_{3} = \begin{pmatrix} 2 & -1 \end{pmatrix} N ;F_{4} = \begin{pmatrix} 3 & -5 \end{pmatrix} N\)
\(R = \begin{pmatrix} -5 & 4 \end{pmatrix} N + \begin{pmatrix} 2 \\ 5 \end{pmatrix} N + \begin{pmatrix} 2 & -1 \end{pmatrix} N + \begin{pmatrix} 3 & -5 \end{pmatrix} N = \begin{pmatrix} 2 \\ 3 \end{pmatrix} N\)
(ii) Let the fifth force be F.
\(F + \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)
\(F = \begin{pmatrix} -2 \\ -3 \end{pmatrix}\)
(b) (i)
Acceleration = a = 2.5 ms\(^{-2}\),
Let retardation be r ms\(^{-2}\).
\(a : r = 3 : 4\)
\(\therefore r = \frac{4}{3} \times 2.5 = 3\frac{1}{3} ms^{-2}\)
(ii) We use the formula \(v = u + at\) for the retardation period.
\(v = 0 m/s ; u = 30 m/s ; a = -3\frac{1}{3} m/s^{2} ; t = ?\)
\(0 = 30 - \frac{10}{3} t \)
\(t = \frac{30 \times 3}{10} = 9 secs\)
(iii) Total distance = area under graph.
\(A_{1} = \frac{1}{2} (4)(20 + 30) = 100m\)
\(A_{2} = 30 \times 8 = 240m\)
\(A_{3} = \frac{1}{2} (9) (30) = 135m\)
Total distance = 100m + 240m + 135m = 475m.
(a) (i) \(F_{1} = \begin{pmatrix} -5 & 4 \end{pmatrix} N; F_{2} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} N; F_{3} = \begin{pmatrix} 2 & -1 \end{pmatrix} N ;F_{4} = \begin{pmatrix} 3 & -5 \end{pmatrix} N\)
\(R = \begin{pmatrix} -5 & 4 \end{pmatrix} N + \begin{pmatrix} 2 \\ 5 \end{pmatrix} N + \begin{pmatrix} 2 & -1 \end{pmatrix} N + \begin{pmatrix} 3 & -5 \end{pmatrix} N = \begin{pmatrix} 2 \\ 3 \end{pmatrix} N\)
(ii) Let the fifth force be F.
\(F + \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)
\(F = \begin{pmatrix} -2 \\ -3 \end{pmatrix}\)
(b) (i)
Acceleration = a = 2.5 ms\(^{-2}\),
Let retardation be r ms\(^{-2}\).
\(a : r = 3 : 4\)
\(\therefore r = \frac{4}{3} \times 2.5 = 3\frac{1}{3} ms^{-2}\)
(ii) We use the formula \(v = u + at\) for the retardation period.
\(v = 0 m/s ; u = 30 m/s ; a = -3\frac{1}{3} m/s^{2} ; t = ?\)
\(0 = 30 - \frac{10}{3} t \)
\(t = \frac{30 \times 3}{10} = 9 secs\)
(iii) Total distance = area under graph.
\(A_{1} = \frac{1}{2} (4)(20 + 30) = 100m\)
\(A_{2} = 30 \times 8 = 240m\)
\(A_{3} = \frac{1}{2} (9) (30) = 135m\)
Total distance = 100m + 240m + 135m = 475m.