# Question 5:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{T+T-30}{2}\right)V$ | M1 | Uses total area under graph to find expression for distance in terms of $V$ and $T$ only. May use trapezium, triangle+rectangle, rectangle minus triangle, or suvat |
| $V(T-15)$ metres | A1* | Given answer correctly obtained (allow omission of 'metres') |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape with acceleration lines parallel and meeting at $(T, V)$ | B1 | B0 if continuous vertical line at $t = T$ |
| Horizontal labels: $(10, 50, 60)$ | B1 | Accept appropriately labelled delineators. Independent of first B1. N.B. If graph not on given graphs, score B0B0 |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{speed}{40} = \frac{V}{30}$ | M1 | Correct method using gradients or suvat to obtain equation in $V$ only |
| $speed = \frac{4V}{3}$ ms$^{-1}$ | A1* | Given answer correctly obtained |

## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\left(\frac{4V}{3}\times 40\right)+\left(\frac{4V}{3}\times 10\right)+\frac{1}{2}\left(\frac{4V}{3}+V\right)(T-60)$ | M1 | Find expression for TOTAL area under motorbike graph (or use suvat) for total distance in terms of $V$ and $T$ only. N.B. omitting the middle section is M0 |
| Correct unsimplified expression with at most one error | A1 | |
| Correct unsimplified expression. Simplified: $\frac{7VT}{6}-30V$ | A1 | |
| Equate motorbike distance to $V(T-15)$ to give equation in $T$ only | M1 | $V$'s must cancel but need not be cancelled for this mark. Independent mark |
| $T = 90$ | A1 | cao |

**ALT method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\left(\frac{V}{3}\right)(T-40+10)$ | A1 | Find area of upper trapezium and parallelogram (differences in areas) |
| $10V$ | A1 | |
| Equate to give equation in $T$ only ($V$ cancels) | M1 | |
| $T=90$ | A1 | |

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