(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) 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 which the acceleration is reduced to 5.0ms\(^{-2}\) for the next 20s. The body maintains the speed attained for 60s after which it is brought to rest in 20s. Draw the velocity-time graph of the motion using the information given above.
(c) Using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest; (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (ii).
(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:
\[
\text{average speed}=\frac{\text{total distance travelled}}{\text{total time taken}}
\]
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 \(20\ \text{s}\).
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 \(340\ \text{m s}^{-1}\).
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 \(20\ \text{s}\).
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 \(340\ \text{m s}^{-1}\).
