a particle moves along a straight line. the graph of the particles position ( x(t) ) at time ( t ) is shown above for ( 0

Question

a particle moves along a straight line. the graph of the particles position ( x(t) ) at time ( t ) is shown above for ( 0 < t < 6 ). the graph has horizontal tangents at ( t = 1 ) and ( t = 5 ) and a point of inflection at ( t = 2 ). for what values of ( t ) is the velocity of the particle increasing?

a ( 0 < t < 2 )
b

Accepted Answer

# Explanation: ## Step1: Recall Velocity and Acceleration Velocity $ v(t) $ is the derivative of position $ x(t) $, so $ v(t)=x'(t) $. Acceleration $ a(t) $ is the derivative of velocity, so $ a(t)=v'(t)=x''(t) $. Velocity increases when acceleration $ a(t)>0 $, which means $ x''(t)>0 $ (concave up of $ x(t) $). ## Step2: Analyze Concavity of $ x(t) $ - A point of inflection is at $ t = 2 $, where concavity changes. - For $ t < 2 $: The graph of $ x(t) $ is concave down (since before $ t = 2 $, the curve bends downward). - For $ t > 2 $: The graph of $ x(t) $ is concave up (after $ t = 2 $, the curve bends upward). Wait, no—wait, let's check the graph. Wait, the position graph: from $ t=0 $ to $ t=2 $, what's the concavity? The point of inflection is at $ t=2 $. Before $ t=2 $, the second derivative (concavity) changes? Wait, no—wait, the graph: at $ t=1 $, horizontal tangent (local min? Wait, at $ t=1 $, the graph has a horizontal tangent, so $ x'(1)=0 $ (velocity zero there). Then from $ t=0 $ to $ t=1 $, the graph is decreasing (velocity negative, since $ x(t) $ goes from 0 to -2, so slope negative). At $ t=1 $, horizontal tangent (slope zero). Then from $ t=1 $ to $ t=5 $, the graph is increasing (slope positive, since it goes from -2 to 1, then to the peak at $ t=5 $ (horizontal tangent there, so $ x'(5)=0 $). Then after $ t=5 $, it decreases a bit. But concavity: the point of inflection is at $ t=2 $. So before $ t=2 $, the graph is concave up or down? Wait, the curve from $ t=0 $ to $ t=2 $: at $ t=0 $, it's at (0,0), goes down to (1, -2) (local min at $ t=1 $), then up to (2, -1) (wait, no, the inflection point is at $ t=2 $). Wait, the inflection point is where the concavity changes. So before $ t=2 $, the second derivative (concavity) is... Let's think about the slope of the velocity (which is acceleration, $ x''(t) $). Velocity is $ x'(t) $, so acceleration is $ x''(t) $, which is the slope of the velocity graph. Alternatively, velocity is $ x'(t) $ (slope of $ x(t) $). To find when velocity is increasing, we need $ x''(t) > 0 $ (since velocity $ v = x' $, so $ v' = x'' $, so $ v $ increases when $ v' > 0 $, i.e., $ x'' > 0 $). So we need to find where $ x(t) $ is concave up (since $ x'' > 0 $ means concave up). Looking at the graph: - From $ t=0 $ to $ t=2 $: Is $ x(t) $ concave up or down? The inflection point is at $ t=2 $. So before $ t=2 $, the graph is concave down (because the curve is bending downward: from $ t=0 $ to $ t=2 $, the slope of $ x(t) $ (velocity) is increasing or decreasing? Wait, at $ t=0 $, the slope of $ x(t) $ is negative (going down from (0,0) to (1, -2)). At $ t=1 $, slope is zero (horizontal tangent). So from $ t=0 $ to $ t=1 $, the slope (velocity) is increasing (from negative to zero: so velocity is increasing, because the slope of $ x(t) $ (velocity) is going from negative to zero, so its derivative (acceleration) is positive? Wait, no—wait, velocity is $ x'(t) $. So if $ x'(t) $ is increasing, then $ x''(t) > 0 $. So when is $ x'(t) $ increasing? Wait, let's track $ x'(t) $ (velocity): - For $ t < 1 $: $ x(t) $ is decreasing (slope negative), and the slope (velocity) is becoming less negative (since at $ t=1 $, slope is zero). So from $ t=0 $ to $ t=1 $, $ x'(t) $ goes from negative to zero: so it's increasing (since it's moving towards zero from negative, so the rate of change of velocity (acceleration) is positive? Wait, no—wait, if $ x'(t) $ is increasing, that means $ x''(t) > 0 $. Let's take an example: if $ x'(t) = -2 $ at $ t=0 $, and $ x'(t) = 0 $ at $ t=1 $, then the change in $ x'(t) $ is $ 0 - (-2) = 2 $ over $ t=0 $ to $ t=1 $, so $ x''(t) $ is positive (since $ x'(t) $ is increasing). - At $ t=1 $ to $ t=2 $: $ x(t) $ is increasing (slope positive, since it goes from -2 to -1, then to 0 at $ t=3 $? Wait, no, the graph crosses the t-axis at $ t=3 $. Wait, maybe I misread the graph. Let's re-express: The position graph: - $ t \in [0,1] $: $ x(t) $ decreases from 0 to -2 (slope negative, velocity negative), reaches a local minimum at $ t=1 $ (horizontal tangent, so $ x'(1)=0 $). - $ t \in [1,5] $: $ x(t) $ increases from -2 to 1 (slope positive, velocity positive), reaches a local maximum at $ t=5 $ (horizontal tangent, so $ x'(5)=0 $). - $ t \in [5,6] $: $ x(t) $ decreases (slope negative, velocity negative). Now, concavity: - Point of inflection at $ t=2 $. So before $ t=2 $, the graph is concave up or down? Wait, the second derivative $ x''(t) $ is the slope of $ x'(t) $ (velocity). So to find when $ x'(t) $ (velocity) is increasing, we need $ x''(t) > 0 $ (since $ x'(t) $ increasing means its derivative is positive). So $ x''(t) > 0 $ when the slope of $ x'(t) $ is positive, i.e., when $ x'(t) $ is increasing. Now, $ x'(t) $ (velocity) has: - At $ t=0 $: slope of $ x(t) $ is negative (since $ x(t) $ is decreasing). - At $ t=1 $: slope of $ x(t) $ is 0 (horizontal tangent). - At $ t=2 $: inflection point, so the slope of $ x'(t) $ (which is $ x''(t) $) changes sign here. - At $ t=5 $: slope of $ x(t) $ is 0 (horizontal tangent). So let's track $ x'(t) $: - From $ t=0 $ to $ t=2 $: Wait, no—wait, the inflection point is at $ t=2 $, so before $ t=2 $, $ x''(t) $ (slope of $ x'(t) $) is positive or negative? Wait, from $ t=0 $ to $ t=1 $: $ x'(t) $ goes from negative to 0 (so it's increasing, so $ x''(t) > 0 $). From $ t=1 $ to $ t=2 $: $ x'(t) $ goes from 0 to positive (since $ x(t) $ is increasing from $ t=1 $ to $ t=5 $), so $ x'(t) $ is increasing (from 0 to positive), so $ x''(t) > 0 $? Wait, no—wait, if $ x'(t) $ is going from 0 to positive, that means it's increasing, so $ x''(t) > 0 $. But at $ t=2 $, inflection point: so after $ t=2 $, does $ x''(t) $ become negative? Wait, no—wait, the inflection point is where concavity changes. So before $ t=2 $, $ x(t) $ is concave up ( $ x''(t) > 0 $ ), and after $ t=2 $, $ x(t) $ is concave down ( $ x''(t) < 0 $ )? Wait, that can't be. Wait, maybe I got the inflection point direction wrong. Wait, the standard: a point of inflection is where the concavity changes. So if the graph of $ x(t) $ has an inflection point at $ t=2 $, then before $ t=2 $, it's concave down, and after $ t=2 $, concave up? Or vice versa. Let's look at the graph: from $ t=0 $ to $ t=2 $, the curve is bending downward (concave down) or upward? Let's take two points: at $ t=0 $, (0,0); $ t=1 $, (1, -2); $ t=2 $, (2, -1). The slope from $ t=0 $ to $ t=1 $ is $ (-2 - 0)/(1 - 0) = -2 $. The slope from $ t=1 $ to $ t=2 $ is $ (-1 - (-2))/(2 - 1) = 1 $. So the slope (velocity) goes from -2 to 1, so it's increasing (from -2 to 1) between $ t=0 $ and $ t=2 $? Wait, no, from $ t=0 $ to $ t=1 $, slope is -2 (at $ t=0 $) to 0 (at $ t=1 $): increasing. From $ t=1 $ to $ t=2 $, slope is 0 (at $ t=1 $) to 1 (at $ t=2 $): increasing. Then from $ t=2 $ to $ t=5 $, the slope (velocity) goes from 1 to 0 (at $ t=5 $, horizontal tangent). So the slope (velocity) is decreasing from $ t=2 $ to $ t=5 $ (since it goes from 1 to 0), so $ x''(t) < 0 $ there. Then after $ t=5 $, slope (velocity) becomes negative (since $ x(t) $ decreases), so $ x'(t) $ goes from 0 to negative, so it's decreasing ( $ x''(t) < 0 $ ). Wait, this is confusing. Let's recall: velocity is $ x'(t) $, acceleration is $ x''(t) $. Velocity increases when $ x''(t) > 0 $ (since acceleration is positive, so velocity is increasing). So when is $ x''(t) > 0 $? - From $ t=0 $ to $ t=2 $: The slope of $ x'(t) $ (which is $ x''(t) $) is positive, because $ x'(t) $ is increasing (from negative to positive, passing through zero at $ t=1 $). - From $ t=2 $ to $ t=5 $: The slope of $ x'(t) $ is negative, because $ x'(t) $ is decreasing (from positive to zero at $ t=5 $). - After $ t=5 $: $ x'(t) $ is decreasing (from zero to negative), so $ x''(t) < 0 $. Wait, but the options are: A. $ 0 < t < 2 $ B. $ 1 < t < 5 $ Wait, maybe I made a mistake. Let's re-express: The key is that velocity $ v(t) = x'(t) $, so $ v(t) $ increasing means $ v'(t) = x''(t) > 0 $ (acceleration positive). So we need to find where $ x(t) $ is concave up (since $ x''(t) > 0 $ means concave up). The graph of $ x(t) $: - Inflection point at $ t=2 $, so before $ t=2 $, $ x(t) $ is concave down? Wait, no—wait, the inflection point is where concavity changes. Let's look at the curve: from $ t=0 $ to $ t=2 $, the curve is concave up (because it's bending upward: the slope of $ x(t) $ is increasing). Wait, when a curve is concave up, the slope (derivative) is increasing. So from $ t=0 $ to $ t=2 $, the slope of $ x(t) $ (velocity) is increasing (from negative to positive), so $ x''(t) > 0 $ (concave up). After $ t=2 $, the slope of $ x(t) $ (velocity) is decreasing (from positive to zero, then to negative), so $ x''(t) < 0 $ (concave down). Wait, but the inflection point is at $ t=2 $, so that's where concavity changes from up to down? No, inflection point is where concavity changes, so if before $ t=2 $ it's concave up, after $ t=2 $ concave down, then $ x''(t) > 0 $ for $ t < 2 $, and $ x''(t) < 0 $ for $ t > 2 $. But the options: A. $ 0 < t < 2 $ B. $ 1 < t < 5 $ Wait, but at $ t=1 $, there's a horizontal tangent ( $ x'(1)=0 $ ), but the concavity: from $ t=0 $ to $ t=2 $, $ x(t) $ is concave up (since slope is increasing), so $ x''(t) > 0 $, so velocity $ v(t) = x'(t) $ is increasing (since $ v'(t) = x''(t) > 0 $). After $ t=2 $, $ x(t) $ is concave down (slope decreasing), so $ x''(t) < 0 $, so velocity is decreasing. Therefore, velocity is increasing when $ x''(t) > 0 $, which is when $ 0 < t < 2 $ (since inflection at $ t=2 $, so before $ t=2 $, concave up, so $ x''(t) > 0 $). Wait, but the option A is $ 0 < t < 2 $, which matches. So the answer should be A. # Answer: A. $ 0 < t < 2 $

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Velocity and Acceleration

Step2: Analyze Concavity of \( x(t) \)

Answer: