A person walks first at a constant speed of 4.90 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 2.70 m/s.
(a) What is her average speed over the entire trip?
The answer is NOT the average of 2.7 and 4.9. I answered this question earlier. Why are you posting it again?
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Vectors and have equal magnitudes of 49.0. If the sum of and is the vector 16.3 , determine the angle between and .
To find the average speed over the entire trip, we need to consider the total distance traveled and the total time taken.
Let's start by calculating the time taken for each leg of the trip.
The distance from point A to point B is the same as the distance from point B to point A since she is moving along a straight line. Therefore, we can just call it d.
The time taken for the first leg of the trip (from A to B) can be calculated using the formula: time = distance / speed.
So, for the first leg, the time taken is t1 = d / 4.90 m/s.
Similarly, for the second leg (from B to A), the time taken is t2 = d / 2.70 m/s.
The total time taken for the entire trip is the sum of the times for the first and second legs: t_total = t1 + t2.
Now, let's calculate the average speed.
The total distance traveled for the entire trip is 2d, as she covers the distance from A to B and back from B to A.
The average speed (v_avg) can be calculated using the formula: average speed = total distance / total time.
Therefore, v_avg = 2d / t_total.
To simplify this expression, we substitute the values of t1 and t2:
v_avg = 2d / (d / 4.90 m/s + d / 2.70 m/s).
We can now simplify further by combining the fractions:
v_avg = 2d / [(d * 2.70 + d * 4.90) / (4.90 * 2.70)].
v_avg = 2d / [(7.20d) / (13.23)].
v_avg = (13.23 * 2d) / (7.20d).
The distance d cancels out, simplifying the expression further:
v_avg = 26.46 / 7.20.
Therefore, the average speed over the entire trip is approximately 3.67 m/s.