MATH SOLVE

4 months ago

Q:
# Refer to the graph and information. An ant is crawling on a straight wire. The velocity, v(t), of the ant at time 0 ≤ t ≤ 8 is given in the graph. Note: The graph on 0 ≤ t ≤ 2 is a semicircle.Part A.Given the velocity curve, at which time (t) is the speed of the ant greatest?A. 0B. 2C. 3D. 4E. 8Part B.An ant’s position during an 8 second time interval is shown by the graph. What is the total distance the ant travelled over the time interval 2 ≤ t ≤ 8?A. 2B. 4C. 6D. 7E. 8Part C.What is the total distance traveled by the ant over the time interval 0 ≤ t ≤ 8?A. 4 - pi/2B. 4 + pi/2C. 4 + piD. 3pi/4 - 5E. 7 + pi/2

Accepted Solution

A:

Part A

Answer: D) 4

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The graph given to use is the velocity of the ant at time t. Look at the graph and locate the highest point, which is at (x,y) = (4,3) where the triangular portion peaks out. They want the x coordinate or the t coordinate, so x = t = 4

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Part B

Answer: D) 7

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Side Note: I'm going to assume there's a typo and it meant to say the ant's velocity rather than position. Otherwise, it would be inconsistent with what part A is saying.

To find the total distance, we find the area under the velocity curve. Any areas below the x axis are treated as positive areas because we want the total distance (instead of the displacement)

The area of the first triangle on the left is 6 square units because

A = b*h/2

A = 4*3/2

A = 12/2

A = 6

The smaller triangle is 1 square unit since

A = b*h/2

A = 2*1/2

A = 2/2

A = 1

The total area of the two triangles combined is 6+1 = 7 square units

The total distance traveled from t = 2 to t = 8 is 7

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Part C

Answer: E) 7+pi/2

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Note: Again I'm assuming that this curve is the velocity (not position) curve

This is essentially identical to part B above. The only difference is the fact that we start at t = 0 this time instead of t = 2

So the answer to part B will be added onto the area of the semicircle

Area of semicircle = (area of full circle)/2

Area of semicircle = (pi*r^2)/2

Area of semicircle = (pi*1^2)/2

Area of semicircle = pi/2

Therefore, we add on pi/2 to the previous answer (back in part B) of 7

That's why the final answer here is 7+(pi/2) or simply 7+pi/2

Answer: D) 4

--------------------------

The graph given to use is the velocity of the ant at time t. Look at the graph and locate the highest point, which is at (x,y) = (4,3) where the triangular portion peaks out. They want the x coordinate or the t coordinate, so x = t = 4

================================================

Part B

Answer: D) 7

--------------------------

Side Note: I'm going to assume there's a typo and it meant to say the ant's velocity rather than position. Otherwise, it would be inconsistent with what part A is saying.

To find the total distance, we find the area under the velocity curve. Any areas below the x axis are treated as positive areas because we want the total distance (instead of the displacement)

The area of the first triangle on the left is 6 square units because

A = b*h/2

A = 4*3/2

A = 12/2

A = 6

The smaller triangle is 1 square unit since

A = b*h/2

A = 2*1/2

A = 2/2

A = 1

The total area of the two triangles combined is 6+1 = 7 square units

The total distance traveled from t = 2 to t = 8 is 7

================================================

Part C

Answer: E) 7+pi/2

--------------------------

Note: Again I'm assuming that this curve is the velocity (not position) curve

This is essentially identical to part B above. The only difference is the fact that we start at t = 0 this time instead of t = 2

So the answer to part B will be added onto the area of the semicircle

Area of semicircle = (area of full circle)/2

Area of semicircle = (pi*r^2)/2

Area of semicircle = (pi*1^2)/2

Area of semicircle = pi/2

Therefore, we add on pi/2 to the previous answer (back in part B) of 7

That's why the final answer here is 7+(pi/2) or simply 7+pi/2