Moderate -0.3 This is a straightforward integration problem requiring students to integrate the acceleration function to find velocity, then substitute t=1.2. While it involves an exponential function, the integration is standard (e^{-0.5t}) and only requires applying the chain rule in reverse plus adding the initial condition. It's slightly easier than average due to being a single-step process with no conceptual challenges beyond basic calculus.
A particle \(P\) moves in a straight line and passes through a point \(O\) of the line with velocity \(2\text{ m s}^{-1}\). At time \(t\) s after passing through \(O\), the velocity of \(P\) is \(v\text{ m s}^{-1}\) and the acceleration of \(P\) is given by \(\text{e}^{-0.5t}\text{ m s}^{-2}\). Calculate the velocity of \(P\) when \(t = 1.2\). [4]
Question 1:
1 | dv
=e−0.5v
dt
1
∫ dv=∫dt
e−0.5v
e–0.5v
=t(+c)
0.5
t = 0, v = 2 so c = 2e
v = 2.4(0) when t = 1.2 | M1
A1
M1
A1 | 4 | Separates the variables and attempts
to integrate
c = 5.4365… or use of limits
2 (i)
(ii)
(iii) | 20(0.8sinθ)
T = AG
0.8
No friction (so perpendicular) AG
20sinθ(0.8cosθ) =
8(0.6sinθ) + 2(1.2sinθ)
θ = 63.3° | B1
B1
M1
A1
A1 | 1
1
3 | Hence 20sinθ
Or ring smooth
Moments about A (3 terms)
All terms correct
Accept 1.1 radians
A particle $P$ moves in a straight line and passes through a point $O$ of the line with velocity $2\text{ m s}^{-1}$. At time $t$ s after passing through $O$, the velocity of $P$ is $v\text{ m s}^{-1}$ and the acceleration of $P$ is given by $\text{e}^{-0.5t}\text{ m s}^{-2}$. Calculate the velocity of $P$ when $t = 1.2$. [4]
\hfill \mbox{\textit{CAIE M2 2015 Q1 [4]}}