(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)

Uniform acceleration is acceleration for which the velocity changes by equal amounts in equal intervals of time. Thus, the acceleration is constant.

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) The successive velocity-time coordinates are:

Plot the points and join successive points with straight lines:

(c)

(i) Maximum speed

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

(ii) Average retardation while the body is being brought to rest

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

Hence, the average retardation is \(\boxed{17\ \text{m s}^{-2}}\).

(iii) Total distance travelled during the first 50 s

The distance is the area under the velocity-time graph from \(0\) to \(50\) s:

\[\begin{aligned} s&=\frac{1}{2}(30)(240)+\frac{1}{2}(240+340)(20)\\&=3600+5800\\&=\boxed{9400\ \text{m}}.\end{aligned}\]

(iv) Average speed during the interval in (ii)

During the 20 s retardation interval, the distance travelled is the triangular area:

\[s=\frac{1}{2}(20)(340)=3400\ \text{m}.\]

\[\text{Average speed}=\frac{3400}{20}=\boxed{170\ \text{m s}^{-1}}.\]