Moderate -0.5 This is a straightforward application of Newton's second law and integration. Given F = ma, students find acceleration a = (3t-4)/0.5, integrate once to get velocity (using v=0 at t=0), then substitute t=4. It's slightly easier than average because it's a direct single-method problem with clear steps and no conceptual complications.
A particle of mass \(0.5\) kg moves in a straight line under the action of a variable force. At time \(t\) seconds, the force is \((3t - 2)\) N in the direction of motion.
Given that the particle starts from rest, find the velocity of the particle when \(t = 4\). [5]
Question 4:
4 | For s = 4.05
Total distance = 4.05 + (3.15 + 4.05)
= 11.25m
tupwards = 0.9
For downwards motion
1
(3.15 + 4.05) = gt 2 → t = 1.2
2
Time taken is 2.1s | M1
A1
B1
B1
B1
B1 | [6] | 2
For using 0 = u – 2gs for the upwards
motion
Page 5 | Mark Scheme | Syllabus | Paper
GCE A LEVEL – May/June 2014 | 9709 | 41
Alternative Mark Scheme for final 3
marks
1
2
[–3.15 = 9T + (–g) T ]
2
2
[100t – 180t – 63 = 0]
(10T – 21)(10T + 3) = 0 | M1
M1
A1 | 1
2
For using s = ut + at for the total
2
displacement and time
For solving a quadratic equation for the
total time T
T = 2.1 only
A particle of mass $0.5$ kg moves in a straight line under the action of a variable force. At time $t$ seconds, the force is $(3t - 2)$ N in the direction of motion.
Given that the particle starts from rest, find the velocity of the particle when $t = 4$. [5]
\hfill \mbox{\textit{CAIE M1 2014 Q4 [5]}}