2. maximum mark: 11
a particle moves along a straight line. its displacement, ( s ) metres, at time ( t ) seconds is given by ( s = t + cos 2t ), ( t geq 0 ). the first two times when the particle is at rest are denoted by ( t_1 ) and ( t_2 ), where ( t_1

Question

2. maximum mark: 11
a particle moves along a straight line. its displacement, ( s ) metres, at time ( t ) seconds is given by ( s = t + cos 2t ), ( t geq 0 ). the first two times when the particle is at rest are denoted by ( t_1 ) and ( t_2 ), where ( t_1 < t_2 ).
(a) find ( t_1 ) and ( t_2 ). 4
(b) find the displacement of the particle when ( t = t_1 ). 2
(c) find the initial velocity of the particle.
(d) find the value of ( t ) when the particle first reaches its maximum velocity.

Accepted Answer

### Part (a) # Explanation: ## Step1: Find velocity function Velocity $ v $ is the derivative of displacement $ s $. Given $ s = t + \cos 2t $, differentiate with respect to $ t $. The derivative of $ t $ is $ 1 $, and the derivative of $ \cos 2t $ is $ -2\sin 2t $ (using chain rule). So $ v = \frac{ds}{dt} = 1 - 2\sin 2t $. ## Step2: Set velocity to zero (at rest) A particle is at rest when $ v = 0 $, so $ 1 - 2\sin 2t = 0 $. Rearrange: $ \sin 2t = \frac{1}{2} $. ## Step3: Solve for $ t $ The general solution for $ \sin \theta = \frac{1}{2} $ is $ \theta = \frac{\pi}{6} + 2k\pi $ or $ \theta = \frac{5\pi}{6} + 2k\pi $, $ k \in \mathbb{Z} $. Here $ \theta = 2t $, so: - For $ 2t = \frac{\pi}{6} + 2k\pi $, $ t = \frac{\pi}{12} + k\pi $ - For $ 2t = \frac{5\pi}{6} + 2k\pi $, $ t = \frac{5\pi}{12} + k\pi $ Since $ t \geq 0 $, the first two positive solutions (smallest $ t_1 < t_2 $): - $ k = 0 $ in first equation: $ t_1 = \frac{\pi}{12} $ - $ k = 0 $ in second equation: $ t_2 = \frac{5\pi}{12} $ # Answer: $ t_1 = \frac{\pi}{12} $, $ t_2 = \frac{5\pi}{12} $ ### Part (b) # Explanation: ## Step1: Substitute $ t = t_1 $ into displacement We know $ t_1 = \frac{\pi}{12} $ and $ s = t + \cos 2t $. Substitute $ t = \frac{\pi}{12} $: ## Step2: Calculate $ \cos 2t $ $ 2t = 2 \times \frac{\pi}{12} = \frac{\pi}{6} $, so $ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} $. ## Step3: Compute $ s $ $ s = \frac{\pi}{12} + \cos \frac{\pi}{6} = \frac{\pi}{12} + \frac{\sqrt{3}}{2} $. # Answer: $ \frac{\pi}{12} + \frac{\sqrt{3}}{2} $ (or simplified form) ### Part (c) # Explanation: ## Step1: Initial velocity is at $ t = 0 $ Initial velocity means $ t = 0 $. Use velocity function $ v = 1 - 2\sin 2t $. ## Step2: Substitute $ t = 0 $ $ 2t = 0 $, so $ \sin 0 = 0 $. Then $ v = 1 - 2(0) = 1 $. # Answer: $ 1 $ m/s ### Part (d) # Explanation: ## Step1: Find acceleration (derivative of velocity) Velocity $ v = 1 - 2\sin 2t $, so acceleration $ a = \frac{dv}{dt} = -4\cos 2t $. ## Step2: Maximum velocity occurs when acceleration is zero (and check sign change) To find maximum velocity, set $ a = 0 $: $ -4\cos 2t = 0 \implies \cos 2t = 0 $. ## Step3: Solve for $ t $ General solution for $ \cos \theta = 0 $ is $ \theta = \frac{\pi}{2} + k\pi $, $ k \in \mathbb{Z} $. Here $ \theta = 2t $, so $ 2t = \frac{\pi}{2} + k\pi \implies t = \frac{\pi}{4} + \frac{k\pi}{2} $. ## Step4: Find first positive $ t $ For $ k = 0 $, $ t = \frac{\pi}{4} $. Check the sign of acceleration around $ t = \frac{\pi}{4} $: before $ t = \frac{\pi}{4} $ (e.g., $ t = 0 $), $ a = -4\cos 0 = -4 < 0 $; after $ t = \frac{\pi}{4} $ (e.g., $ t = \frac{\pi}{2} $), $ a = -4\cos \pi = 4 > 0 $. Wait, no—wait, velocity is $ v = 1 - 2\sin 2t $. Wait, maybe I messed up: maximum velocity: we can also analyze $ v(t) = 1 - 2\sin 2t $. The maximum of $ v $ occurs when $ \sin 2t $ is minimized (since $ v = 1 - 2\sin 2t $). The minimum of $ \sin 2t $ is $ -1 $, but wait, no—wait, to find maximum of $ v $, take derivative of $ v $ (acceleration) and set to zero. Wait, earlier: acceleration $ a = -4\cos 2t $. When $ a = 0 $, $ \cos 2t = 0 $, so $ 2t = \frac{\pi}{2} $ (first time, $ k=0 $) gives $ t = \frac{\pi}{4} $. But wait, let's check $ v(t) $ at $ t = \frac{\pi}{4} $: $ v = 1 - 2\sin \frac{\pi}{2} = 1 - 2(1) = -1 $. Wait, that's a minimum? Wait, no, I made a mistake. Wait, velocity is $ v = 1 - 2\sin 2t $. To find maximum of $ v $, we can see that $ \sin 2t $ ranges from -1 to 1. So $ v = 1 - 2\sin 2t $ will have maximum when $ \sin 2t $ is minimum (since it's subtracted). The minimum of $ \sin 2t $ is -1, so $ v_{max} = 1 - 2(-1) = 3 $. When does $ \sin 2t = -1 $? $ 2t = \frac{3\pi}{2} + 2k\pi \implies t = \frac{3\pi}{4} + k\pi $. Wait, this is conflicting with acceleration. Wait, no—my error: acceleration is derivative of velocity, so to find extrema of velocity, set acceleration to zero. But if $ v = 1 - 2\sin 2t $, then $ a = -4\cos 2t $. So when $ a = 0 $, $ \cos 2t = 0 $, so $ 2t = \frac{\pi}{2} $ or $ \frac{3\pi}{2} $, etc. At $ t = \frac{\pi}{4} $, $ 2t = \frac{\pi}{2} $, $ \sin 2t = 1 $, so $ v = 1 - 2(1) = -1 $. At $ t = \frac{3\pi}{4} $, $ 2t = \frac{3\pi}{2} $, $ \sin 2t = -1 $, so $ v = 1 - 2(-1) = 3 $. Now, check acceleration sign: at $ t < \frac{\pi}{4} $ (e.g., $ t = 0 $), $ a = -4\cos 0 = -4 < 0 $, so velocity is decreasing. At $ \frac{\pi}{4} < t < \frac{3\pi}{4} $ (e.g., $ t = \frac{\pi}{2} $), $ a = -4\cos \pi = 4 > 0 $, so velocity is increasing. At $ t > \frac{3\pi}{4} $ (e.g., $ t = \pi $), $ a = -4\cos 2\pi = -4 < 0 $, velocity is decreasing. So the velocity has a minimum at $ t = \frac{\pi}{4} $ and a maximum at $ t = \frac{3\pi}{4} $? Wait, no—wait, the function $ v(t) = 1 - 2\sin 2t $. Let's plot it: period $ \pi $. At $ t = 0 $, $ v = 1 $. At $ t = \frac{\pi}{4} $, $ v = -1 $. At $ t = \frac{\pi}{2} $, $ v = 1 - 2\sin \pi = 1 $. At $ t = \frac{3\pi}{4} $, $ v = 3 $. At $ t = \pi $, $ v = 1 $. So the velocity oscillates between -1 and 3, with period $ \pi $. So the first maximum (after $ t=0 $) is at $ t = \frac{3\pi}{4} $? Wait, no—at $ t=0 $, $ v=1 $. Then it decreases to $ v=-1 $ at $ t=\frac{\pi}{4} $, then increases to $ v=3 $ at $ t=\frac{3\pi}{4} $, then decreases back to $ v=1 $ at $ t=\pi $. So the first time it reaches maximum velocity (3) is at $ t = \frac{3\pi}{4} $? But that contradicts the acceleration method. Wait, no—my mistake in acceleration sign: when $ a > 0 $, velocity is increasing; when $ a < 0 $, velocity is decreasing. So from $ t=0 $ to $ t=\frac{\pi}{4} $, $ a < 0 $ (decreasing), from $ t=\frac{\pi}{4} $ to $ t=\frac{3\pi}{4} $, $ a > 0 $ (increasing), from $ t=\frac{3\pi}{4} $ onwards, $ a < 0 $ (decreasing). So the velocity decreases from $ t=0 $ to $ t=\frac{\pi}{4} $ (reaching minimum at $ t=\frac{\pi}{4} $), then increases from $ t=\frac{\pi}{4} $ to $ t=\frac{3\pi}{4} $ (reaching maximum at $ t=\frac{3\pi}{4} $), then decreases again. So the first time it reaches maximum velocity is at $ t = \frac{3\pi}{4} $? But wait, the problem says "first reaches its maximum velocity". Wait, at $ t=0 $, $ v=1 $. Then it goes down to -1, then up to 3. So the maximum velocity is 3, first achieved at $ t = \frac{3\pi}{4} $. But let's check the acceleration approach again. Wait, maybe I messed up the derivative. Wait, $ s = t + \cos 2t $, so $ v = 1 - 2\sin 2t $ (correct, derivative of $ \cos 2t $ is $ -2\sin 2t $). Then $ a = -4\cos 2t $ (correct, derivative of $ -2\sin 2t $ is $ -4\cos 2t $). So to find extrema of $ v $, set $ a = 0 $, so $ \cos 2t = 0 \implies 2t = \frac{\pi}{2} + k\pi \implies t = \frac{\pi}{4} + \frac{k\pi}{2} $. For $ k=0 $, $ t=\frac{\pi}{4} $ (minimum, since before that $ a < 0 $ (decreasing), after that $ a > 0 $ (increasing)). For $ k=1 $, $ t=\frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} $ (maximum, since before that $ a > 0 $ (increasing), after that $ a < 0 $ (decreasing)). So yes, the first maximum velocity occurs at $ t = \frac{3\pi}{4} $. Wait, but let's confirm with the velocity function. $ v(t) = 1 - 2\sin 2t $. The maximum of $ v $ occurs when $ \sin 2t $ is minimized (because it's $ 1 - 2\sin 2t $). The minimum of $ \sin 2t $ is -1, so $ v_{max} = 1 - 2(-1) = 3 $. When does $ \sin 2t = -1 $? $ 2t = \frac{3\pi}{2} + 2k\pi \implies t = \frac{3\pi}{4} + k\pi $. For $ k=0 $, $ t = \frac{3\pi}{4} $, which matches the acceleration method (since at $ t=\frac{3\pi}{4} $, $ a = -4\cos \frac{3\pi}{2} = 0 $, and before that $ a > 0 $ (increasing), after that $ a < 0 $ (decreasing), so it's a maximum). So the first time the particle reaches maximum velocity is at $ t = \frac{3\pi}{4} $. # Answer: $ \frac{3\pi}{4} $

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Find velocity function

Step2: Set velocity to zero (at rest)

Step3: Solve for \( t \)

Step1: Substitute \( t = t_1 \) into displacement

Step2: Calculate \( \cos 2t \)

Step3: Compute \( s \)

Step1: Initial velocity is at \( t = 0 \)

Step2: Substitute \( t = 0 \)

Answer:

Part (b)