When a stone is thrown vertically upwards, its distance d metres after t seconds is given by the formula \(d = 60t - 10t^{2}\). Draw the graph of \(d = 60t ...
Question 1 Report
When a stone is thrown vertically upwards, its distance d metres after t seconds is given by the formula \(d = 60t - 10t^{2}\). Draw the graph of \(d = 60t - 10t^{2}\) for values of t from 1 to 5 seconds using 2cm to 1 unit on the t- axis and 2cm to 20 units on the d- axis.
(a) Using your graph, (i) how long does it take to reach a height of 70 metres? (ii) determine the height of the stone after 5 seconds. (iii) after how many seconds does it reach its maximum height.
(b) Determine the slope of the graph when t = 4 seconds.
Answer Details
Table of values for \(d = 60t - 10t^2\):
| \(t\) (s) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| \(d\) (m) | 50 | 80 | 90 | 80 | 50 |
Plot these points (2 cm to 1 unit on the \(t\)-axis, 2 cm to 20 units on the \(d\)-axis) and join with a smooth curve (a downward parabola).
(a)(i) Time to reach 70 m: Draw the line \(d = 70\); it cuts the curve twice. Solving \(60t - 10t^2 = 70\) gives \(t^2 - 6t + 7 = 0\), so \(t = 3 \pm \sqrt{2}\), i.e. \(t \approx \mathbf{1.6\text{ s}}\) (rising) and \(t \approx \mathbf{4.4\text{ s}}\) (falling).
(ii) Height after 5 s: from the curve, \(d = \mathbf{50\text{ m}}\).
(iii) Time at maximum height: the curve peaks at the vertex, \(t = \dfrac{60}{2(10)} = \mathbf{3\text{ s}}\) (height 90 m).
(b) Slope at \(t = 4\) s: the gradient of the tangent. Since \(\dfrac{dd}{dt} = 60 - 20t\), at \(t = 4\): \(60 - 20(4) = \mathbf{-20}\) (m/s). A tangent drawn at \(t = 4\) on the graph gives approximately this value.
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