Higher June 2022 Paper 5 Q7
7
(a) A car accelerates at 4.06 m/s2 for 10.1 seconds from an initial velocity of 2.93 m/s.
Harper rounds each value to 1 significant figure.
Harper uses the rounded values and the formula
to estimate the distance travelled in the 10.1 seconds.
Harper’s answer is 430 metres.
Using Harper’s method, show that their answer is wrong. [4]
(b) Rearrange this formula to make \(t\) the subject.\[s = \tfrac{1}{2}at^2\]
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(s = 230\) with 4, 3 and 10 or 100 seen | 4 | B2 for 4, 3 and 10 or 100 or B1 for two correct | For all marks condone e.g. 3.00, 4.0, 10.0 used These values may be written in the stem of the question |
| AND M1 for \((3 \times 10) + \tfrac{1}{2}(4 \times 10^2)\) or correct substitution of unrounded or incorrectly rounded values If 0 scored then SC1 for sight of 230 | For M1 e.g. allow a mixture \((2.93 \times 10.1) + \tfrac{1}{2}(4.1 \times 10^2)\) | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(t = [\pm]\sqrt{\dfrac{2s}{a}}\) oe final answer | 3 | Square root must dip below fraction line in final answer unless \(\left(\frac{2s}{a}\right)\) bracketed For 3 marks oe e.g. \(t = [\pm]\sqrt{\dfrac{2 \times s}{a}}\), \(t = \sqrt{\dfrac{s}{\frac{1}{2}a}}\) | |
| M2 for first two steps correctly completed e.g. \(\frac{2s}{a} = t^2\) or answer \([\pm]\sqrt{\dfrac{2s}{a}}\) (no \(t =\) ) | M2 for e.g. \(\dfrac{s}{\frac{a}{2}} = t^2\), \(\dfrac{s}{\frac{1}{2}a} = t^2\), \(\dfrac{s}{0.5a} = t^2\) | ||
| or M1 for first step correctly completed e.g. \(2s = at^2\) or \(\frac{s}{a} = \frac{1}{2}t^2\) | Allow M1 for \(\frac{s}{0.5} = at^2\) oe | ||
| If 0 scored, SC1 for final answer \(t = [\pm]\sqrt{\dfrac{\frac{1}{2}s}{a}}\) oe | oe for SC1 e.g. \(t = [\pm]\sqrt{\dfrac{s}{2a}}\) | ||