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Therefore \(X = 1.3 - 0.1 = 1.20\).
This is a linear ODE with $P = 3\sec^2 x$ and $Q = \sec^2 x$.
Integrating factor: $e^{\int 3\sec^2 x\,dx} = e^{3\tan x}$
$\frac{d}{dx}(ye^{3\tan x}) = \sec^2 x \cdot e^{3\tan x}$
Integrate: $ye^{3\tan x} = \frac{1}{3}e^{3\tan x} + C$
At $x=\pi/4$, $\tan(\pi/4)=1$, $y=4/3$: $\frac{4}{3}e^3 = \frac{1}{3}e^3 + C \Rightarrow C = e^3$
So $y = \frac{1}{3} + e^3 \cdot e^{-3\tan x}$
At $x=-\pi/4$, $\tan(-\pi/4)=-1$: $y = \frac{1}{3} + e^3 \cdot e^{3} = \frac{1}{3} + e^3$... wait: $e^{-3(-1)}=e^3$, so $y = \frac{1}{3}+e^3 \cdot e^3 = \frac{1}{3}+e^6$.
Correct: $y(-\pi/4) = \mathbf{\frac{1}{3}+e^6}$. So correct option index is 0.
Using $\vec{s} = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2$:
- $x = 2 + 5(2) + \frac{1}{2}(4)(2^2) = 2 + 10 + 8 = 20$
- $y = 4 + 4(2) + \frac{1}{2}(4)(2^2) = 4 + 8 + 8 = 20$
- Distance from origin $D = \sqrt{20^2 + 20^2} = 20\sqrt{2}\text{ m}$.
The coordinates of a particle at any time $t$ are $x = 5t - 2t^2$ and $y = 10t$ (metres, seconds). The acceleration of the particle at $t = 2\text{ s}$ is:
$a_x = \dfrac{d^2x}{dt^2} = -4\text{ m/s}^2$ (constant)
$a_y = \dfrac{d^2y}{dt^2} = 0$
Acceleration vector $= -4\hat{i}\text{ m/s}^2$ at all times, including $t = 2\text{ s}$. Magnitude $= \mathbf{4\text{ m/s}^2}$, directed in $-x$ direction.
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