| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with all parameters given |
| Difficulty | Moderate -0.8 This is a straightforward two-stage kinematics problem using standard SUVAT equations with clearly stated accelerations and times. Part (a) requires one application of v=u+at, and part (b) requires calculating distances for two separate constant acceleration phases then summing—routine mechanics with no conceptual challenges or problem-solving insight needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks |
|---|---|
| (d) | Speed ms-1 |
| Answer | Marks |
|---|---|
| V = 50.4 | M1A1 |
Question 2:
2
4. (a)
(b)
(c)
(d) | Speed ms-1
36
Shape
Time (s)
90
36,90
Time to accelerate : time to decelerate
18 s 12 s
Distance= area under graph
1
= (cid:4) 36 (cid:4) (90 + 120) m
2
= 3780 m
There is no period of constant maximum velocity
(OR “it speeds up and then immediately slows down again” OR “it attains a
greater maximum speed”)
Let greatest speed be V m s−1 then
1
(cid:4) 150 (cid:4) V = 3780
max
2
V = 50.4 | M1A1
B1 (3)
M1 A1
M1 A1
A1 (5)
B1 (1)
M1 A1
A1
(9 marks)A car starts from rest at a point $O$ and moves in a straight line. The car moves with constant acceleration $4 \text{ m s}^{-2}$ until it passes the point $A$ when it is moving with speed $10 \text{ m s}^{-1}$. It then moves with constant acceleration $3 \text{ m s}^{-2}$ for 6 s until it reaches the point $B$. Find
\begin{enumerate}[label=(\alph*)]
\item the speed of the car at $B$, [2]
\item the distance $OB$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q2 [7]}}