A body, moving at 20ms\(^{-1}\) accelerates... | WAEC SSCE - Further Mathematics - 2019 (Essay)

Question 1 Report

A body, moving at 20ms\(^{-1}\) accelerates uniformly at 2\(\frac{1}{2}ms^{-2}\) for 4 seconds. It continues the journey at this speed for 8 seconds, before coming to rest seconds at tseconds after with uniform retardation. If the ratio of the acceleration to retardation is 3 : 4

(a( sketch the velocity - times graph of the journey

(b) find t

Answer

(a) Velocity-time graph

The velocity increases uniformly from \(20\text{ m s}^{-1}\) to \(30\text{ m s}^{-1}\) in 4 s, remains constant for 8 s, and then decreases uniformly to zero.

(b) Calculation of \(t\)

Acceleration \(=2\frac{1}{2}=\frac{5}{2}\text{ m s}^{-2}\).

Let the retardation be \(r\text{ m s}^{-2}\). Since

\[\frac{5}{2}:r=3:4,\]

\[r=\frac{4}{3}\times\frac{5}{2}=\frac{10}{3}\text{ m s}^{-2}.\]

Velocity after the first 4 s is

\[v=20+\left(\frac{5}{2}\times4\right)=30\text{ m s}^{-1}.\]

During retardation,

\[0=30-\frac{10}{3}t.\]

\[t=\frac{30}{10/3}=9\text{ s}.\]

Thus, the body comes to rest at \(4+8+9=21\) s from the start of the journey.

(c) Total distance travelled

The total distance is the area under the velocity-time graph:

\[\begin{aligned} \text{Distance}&=\frac{1}{2}(20+30)(4)+(30\times8)+\frac{1}{2}(30)(9)\\ &=100+240+135\\ &=475\text{ m}. \end{aligned}\]

Total distance travelled = \(475\text{ m}\).

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