(a) Explain the terms:- uniform acceleration and average speed. (b) A body at rest is given an initial uniform acceleration of 8.0ms\(^{-2}\) for 30s after ...

(a) Definitions

Uniform acceleration means that velocity changes by equal amounts in equal time intervals. Therefore, the acceleration is constant.

\[ a=\frac{v-u}{t} \]

Average speed is the total distance travelled divided by the total time taken:

\[ \text{average speed}=\frac{\text{total distance travelled}}{\text{total time taken}} \]

(b) Velocity–time graph

Calculate the speeds at the end of each stage:

\[ v=0+(8.0)(30)=240\ \text{m s}^{-1} \] \[ v=240+(5.0)(20)=340\ \text{m s}^{-1} \]

The body then travels at \(340\ \text{m s}^{-1}\) for \(60\ \text{s}\), from \(t=50\ \text{s}\) to \(t=110\ \text{s}\), before slowing uniformly to rest by \(t=130\ \text{s}\).

(c)(i) Maximum speed

The maximum speed is the highest value on the graph:

\[ \boxed{340\ \text{m s}^{-1}} \]

(c)(ii) Average retardation while coming to rest

The body slows from \(340\ \text{m s}^{-1}\) to \(0\ \text{m s}^{-1}\) in \(20\ \text{s}\).

\[ a=\frac{v-u}{t}=\frac{0-340}{20}=-17\ \text{m s}^{-2} \]

The acceleration is negative because the velocity is decreasing. The retardation, stated as a positive magnitude, is:

\[ \boxed{17\ \text{m s}^{-2}} \]

(c)(iii) Total distance travelled during the first \(50\ \text{s}\)

Distance is the area under a velocity–time graph.

From \(0\) to \(30\ \text{s}\), the area is a triangle:

\[ s_1=\frac{1}{2}\times30\times240=3600\ \text{m} \]

From \(30\) to \(50\ \text{s}\), the area is a trapezium:

\[ s_2=\frac{1}{2}(240+340)\times20=5800\ \text{m} \] \[ \text{Total distance}=3600+5800=\boxed{9400\ \text{m}} \]

(c)(iv) Average speed during the first \(50\ \text{s}\)

\[ \text{average speed}=\frac{9400}{50}=\boxed{188\ \text{m s}^{-1}} \]

Important correction: The values \(240\ \text{m s}^{-1}\), \(200\ \text{m s}^{-1}\), and \(160\ \text{m s}^{-1}\) in the supplied reference answer do not follow from the stated motion. After reaching \(240\ \text{m s}^{-1}\) at \(30\ \text{s}\), the body continues to accelerate at \(5.0\ \text{m s}^{-2}\) for \(20\ \text{s}\), so its speed must increase by \(100\ \text{m s}^{-1}\) to \(340\ \text{m s}^{-1}\).