Moderate -0.8 This is a straightforward application of the area-under-graph principle for displacement, requiring students to calculate the area of a trapezium and solve a simple linear equation. It's below average difficulty as it tests only basic kinematics interpretation with no problem-solving complexity.
Fig. 1 is the velocity-time graph for the motion of a body. The velocity of the body is \(v\text{ m s}^{-1}\) at
time \(t\) seconds.
\includegraphics{figure_1}
The displacement of the body from \(t = 0\) to \(t = 100\) is 1400 m. Find the value of \(V\). [4]
Attempt to find areas in terms of ratios (at least
one
correct)
Correct total ratio – need not be evaluated.
(Evidence
may be 800 or 400 or 200 seen).
Complete method. (Evidence may be 800/40 or
400/20
or 200/10 seen).
cao
[ Award 3/4 for 20 seen WWW]
4
Answer
Marks
mark
Sub
Question 2:
2 | either
70V obtained
So 70V = 1400
and V = 20
or
V = 20 | M1
A1
M1
A1
M1
A1
M1
A1 | Attempt at area. If not trapezium method at least
one
part area correct. Accept equivalent.
Or equivalent – need not be evaluated.
Equate their 70V to 1400. Must have attempt at
complete areas or equations.
cao
Attempt to find areas in terms of ratios (at least
one
correct)
Correct total ratio – need not be evaluated.
(Evidence
may be 800 or 400 or 200 seen).
Complete method. (Evidence may be 800/40 or
400/20
or 200/10 seen).
cao
[ Award 3/4 for 20 seen WWW]
4
mark | Sub
Fig. 1 is the velocity-time graph for the motion of a body. The velocity of the body is $v\text{ m s}^{-1}$ at
time $t$ seconds.
\includegraphics{figure_1}
The displacement of the body from $t = 0$ to $t = 100$ is 1400 m. Find the value of $V$. [4]
\hfill \mbox{\textit{OCR MEI M1 Q2 [4]}}