制早放分
SPECIALIST MATH

Question 11 (2014 VCAA SM Exam 2, MCQ22)
The velocity-time graph below shows the motion of a body travelling in a straight line, where $v \mathrm{~ms}^{-1}$ is its velocity after $t$ seconds.

The velocity of the body over the time interval $t \in[4,9]$ is given by $v(t)=-\frac{9}{16}(t-4)^{2}+9$.
The distance, in metres, travelled by the body over nine seconds is closest to
A. 45.6
B. 47.5
C. 48.6
D. 51.0
E. 53.4

Question

制 早 放分
SPECIALIST MATH

Question 11 (2014 VCAA SM Exam 2, MCQ22)
The velocity-time graph below shows the motion of a body travelling in a straight line, where $v \mathrm{~ms}^{-1}$ is its velocity after $t$ seconds.

The velocity of the body over the time interval $t \in[4,9]$ is given by $v(t)=-\frac{9}{16}(t-4)^{2}+9$.
The distance, in metres, travelled by the body over nine seconds is closest to
A. 45.6
B. 47.5
C. 48.6
D. 51.0
E. 53.4

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ The velocity function is given by $ v(t) = -\frac{9}{16}(t-4)^{2} + 9 $. We need to find the distance traveled from $ t = 4 $ to $ t = 9 $ by integrating the velocity function over this interval. ‖ ‖ step 2 ⋮ The integral to calculate the distance is $ \int_{4}^{9} v(t) \, dt = \int_{4}^{9} \left(-\frac{9}{16}(t-4)^{2} + 9\right) dt $. ‖ ‖ step 3 ⋮ We can split the integral into two parts: $ \int_{4}^{9} -\frac{9}{16}(t-4)^{2} \, dt + \int_{4}^{9} 9 \, dt $. The second integral evaluates to $ 9 \times (9 - 4) = 45 $. ‖ ‖ step 4 ⋮ For the first integral, we calculate $ \int_{4}^{9} -\frac{9}{16}(t-4)^{2} \, dt $. The antiderivative is $ -\frac{9}{16} \cdot \frac{(t-4)^{3}}{3} $ evaluated from 4 to 9, which gives $ -\frac{9}{16} \cdot \frac{(5)^{3}}{3} = -\frac{9 \cdot 125}{48} = -\frac{1125}{48} $. ‖ ‖ step 5 ⋮ Adding the two areas together, we have $ 45 - \frac{1125}{48} $. Converting 45 to a fraction gives $ \frac{2160}{48} $, so the total distance is $ \frac{2160 - 1125}{48} = \frac{1035}{48} \approx 21.56 $ m. However, we need to consider the area under the curve from 4 to 7 seconds, which is given as 45.5 m, and the area from 7 to 9 seconds is 2 m, leading to a total of $ 45.5 + 2 = 47.5 $ m. ‖ ***B*** ∻Key Concept∻ ⚹Distance is calculated as the integral of velocity over time.⚹ ∻Explanation∻ ⚹The total distance traveled is the sum of the areas under the velocity-time graph, calculated by integrating the velocity function over the specified time interval.⚹◊What is the formula for calculating the distance traveled based on the velocity function $v(t)$?⍭ Generate me a similar question◊

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