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Sia

Question
Math
Posted 10 months ago
Scarlett received the following problem:
A particle moves in a straight line with velocity v(t)=2t+5v(t)=2 t+5 meters per second, where tt is time in seconds. At t=1t=1, the particle's distance from the starting point was 5 meters in the positive direction. What is the particle's displacement between t=1t=1 to t=3t=3 seconds?

Which expression should Scarlett use to solve the problem?
Choose 1 answer:
(A) 13v(t)dt+5\int_{1}^{3} v(t) d t+5
(B) 13v(t)dt\int_{1}^{3} v(t) d t
(C) v(3)v(1)v(3)-v(1)
(D) v(3)+5v^{\prime}(3)+5
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Answer from Sia
Posted 10 months ago
Solution by Steps
step 1
To find the particle's displacement between t=1t=1 and t=3t=3 seconds, we need to integrate the velocity function v(t)=2t+5v(t)=2t+5 over the interval from t=1t=1 to t=3t=3
step 2
The integral of the velocity function will give us the change in position, or displacement, from t=1t=1 to t=3t=3
step 3
The displacement is given by the definite integral 13v(t)dt\int_{1}^{3} v(t) dt
step 4
Since the particle's distance from the starting point at t=1t=1 was 5 meters, we add this initial distance to the result of the integral to find the total displacement
step 5
Therefore, the expression Scarlett should use to solve the problem is 13v(t)dt+5\int_{1}^{3} v(t) dt + 5
Answer
(A) 13v(t)dt+5\int_{1}^{3} v(t) dt+5
Key Concept
Displacement from Velocity Function
Explanation
The displacement of a particle is found by integrating its velocity function over the given time interval and adding the initial position if provided.

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