Physics 151 Class Exercise: Vectors2 - KEY 1. A basketball player runs down the court, following the path indicated by the vectors A, B, and C in the figure below. The magnitudes of these three vectors are A = 10.0 m, B = 20.0 m, and C = 7.0 m. Find the magnitude and direction of the net displacement of the player.
On occasion objects move within a medium that is moving with respect to an observer. For example, an airplane usually encounters a wind - air that is moving with respect to an observer on the ground below. As another example, a motorboat in a river is moving amidst a river current - water that is moving with respect to an observer on dry land. In such instances as this, the magnitude of the velocity of the moving object (whether it be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. That is to say, the speedometer on the motorboat might read 20 mi/hr; yet the motorboat might be moving relative to the observer on shore at a speed of 25 mi/hr. Motion is relative to the observer. The observer on land, often named (or misnamed) the 'stationary observer' would measure the speed to be different than that of the person on the boat.
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The observed speed of the boat must always be described relative to who the observer is.Tailwinds, Headwinds, and Side WindsTo illustrate this principle, consider a plane flying amidst a tailwind. A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity. If the plane is traveling at a velocity of 100 km/hr with respect to the air, and if the wind velocity is 25 km/hr, then what is the velocity of the plane relative to an observer on the ground below?
The resultant velocity of the plane (that is, the result of the wind velocity contributing to the velocity due to the plane's motor) is the vector sum of the velocity of the plane and the velocity of the wind. This resultant velocity is quite easily determined if the wind approaches the plane directly from behind. As shown in the diagram below, the plane travels with a resulting velocity of 125 km/hr relative to the ground.If the plane encounters a headwind, the resulting velocity will be less than 100 km/hr. Since a headwind is a wind that approaches the plane from the front, such a wind would decrease the plane's resulting velocity. Suppose a plane traveling with a velocity of 100 km/hr with respect to the air meets a headwind with a velocity of 25 km/hr. In this case, the resultant velocity would be 75 km/hr; this is the velocity of the plane relative to an observer on the ground.
This is depicted in the diagram below.Now consider a plane traveling with a velocity of 100 km/hr, South that encounters a side wind of 25 km/hr, West. Now what would the resulting velocity of the plane be?
This question can be answered in the same manner as the previous questions. The resulting velocity of the plane is the vector sum of the two individual velocities.
To determine the resultant velocity, the plane velocity (relative to the air) must be added to the wind velocity. This is the same procedure that was used above for the headwind and the tailwind situations; only now, the resultant is not as easily computed. Since the two vectors to be added - the southward plane velocity and the westward wind velocity - are at right angles to each other, the can be used.
This is illustrated in the diagram below.In this situation of a side wind, the southward vector can be added to the westward vector using the. The magnitude of the resultant velocity is determined using Pythagorean theorem.
The algebraic steps are as follows: (100 km/hr) 2 + (25 km/hr) 2 = R 210 000 km 2/hr 2 + 625 km 2/hr 2 = R 210 625 km 2/hr 2 = R 2SQRT(10 625 km 2/hr 2) = R103.1 km/hr = RThe direction of the resulting velocity can be determined using a. Since the plane velocity and the wind velocity form a right triangle when added together in head-to-tail fashion, the angle between the resultant vector and the southward vector can be determined using the sine, cosine, or tangent functions.
The tangent function can be used; this is shown below:tan (theta) = (opposite/adjacent)tan (theta) = (25/100)theta = invtan (25/100)theta = 14.0 degreesIf the resultant velocity of the plane makes a 14.0 degree angle with the southward direction (theta in the above diagram), then the direction of the resultant is 256 degrees. Like any vector, the resultant's is measured as a counterclockwise angle of rotation from due East. Analysis of a Riverboat's MotionThe effect of the wind upon the plane is similar to the effect of the river current upon the motorboat.
If a motorboat were to head straight across a river (that is, if the boat were to point its bow straight towards the other side), it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream. The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction. While the speedometer of the boat may read 4 m/s, its speed with respect to an observer on the shore will be greater than 4 m/s.The resultant velocity of the motorboat can be determined in the same manner as was done for the plane.
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The resultant velocity of the boat is the vector sum of the boat velocity and the river velocity. Since the boat heads straight across the river and since the current is always directed straight downstream, the two vectors are at right angles to each other.
Thus, the can be used to determine the resultant velocity. Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. What would be the resultant velocity of the motorboat (i.e., the velocity relative to an observer on the shore)? The magnitude of the resultant can be found as follows: (4.0 m/s) 2 + (3.0 m/s) 2 = R 216 m 2/s 2 + 9 m 2/s 2 = R 225 m 2/s 2 = R 2SQRT (25 m 2/s 2) = R5.0 m/s = RThe of the resultant is the counterclockwise angle of rotation that the resultant vector makes with due East. This angle can be determined using a trigonometric function as shown below. The resultant velocity can be found using the Pythagorean theorem. The resultant is the hypotenuse of a right triangle with sides of 4 m/s and 7 m/s.
It isSQRT (4 m/s) 2 + (7 m/s) 2 = 8.06 m/sIts direction can be determined using a trigonometric function.Direction = invtan (7 m/s) / (4 m/s) = 60°b. The time to cross the river is t = d / v = (80 m) / (4 m/s) = 20 sc. The distance traveled downstream is d = v. t = (7 m/s). (20 s) = 140 mAn important concept emerges from the analysis of the two example problems above. In Example 1, the time to cross the 80-meter wide river (when moving 4 m/s) was 20 seconds.
This was in the presence of a 3 m/s current velocity. In Example 2, the current velocity was much greater - 7 m/s - yet the time to cross the river remained unchanged. In fact, the current velocity itself has no effect upon the time required for a boat to cross the river. The river moves downstream parallel to the banks of the river.
As such, there is no way that the current is capable of assisting a boat in crossing a river. While the increased current may affect the resultant velocity - making the boat travel with a greater speed with respect to an observer on the ground - it does not increase the speed in the direction across the river. The component of the resultant velocity that is increased is the component that is in a direction pointing down the river.
It is often said that 'perpendicular components of motion are independent of each other.' As applied to riverboat problems, this would mean that an across-the-river variable would be independent of (i.e., not be affected by) a downstream variable. The time to cross the river is dependent upon the velocity at which the boat crosses the river. It is only the component of motion directed across the river (i.e., the boat velocity) that affects the time to travel the distance directly across the river (80 m in this case). The component of motion perpendicular to this direction - the current velocity - only affects the distance that the boat travels down the river.
This concept of perpendicular components of motion will be investigated in more detail in the.Check Your Understanding1. A plane can travel with a speed of 80 mi/hr with respect to the air.
Determine the resultant velocity of the plane (magnitude only) if it encounters a a. 10 mi/hr headwind.b. 10 mi/hr tailwind.c. 10 mi/hr crosswind.d. 60 mi/hr crosswind. A headwind would decrease the resultant velocity of the plane to 70 mi/hr.b.
A tailwind would increase the resultant velocity of the plane to 90 mi/hr.c. A 10 mi/hr crosswind would increase the resultant velocity of the plane to 80.6 mi/hr.This can be determined using the Pythagorean theorem: SQRT (80 mi/hr) 2 + (10 mi/hr) 2 )d. A 60 mi/hr crosswind would increase the resultant velocity of the plane to 100 mi/hr.This can be determined using the Pythagorean theorem: SQRT (80 mi/hr) 2 + (60 mi/hr) 2 )2. A motorboat traveling 5 m/s, East encounters a current traveling 2.5 m/s, North.
What is the resultant velocity of the motor boat?b. If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?c. What distance downstream does the boat reach the opposite shore? The resultant velocity can be found using the Pythagorean theorem. The resultant is the hypotenuse of a right triangle with sides of 5 m/s and 2.5 m/s. It isSQRT(5 m/s) 2 + (2.5 m/s) 2 = 5.59 m/sIts direction can be determined using a trigonometric function.Direction = invtan (2.5 m/s) / (5 m/s) = 26.6 degreesb.
The time to cross the river is t = d / v = (80 m) / (5 m/s) = 16.0 sc. The distance traveled downstream is d = v. t = (2.5 m/s).(16.0 s) = 40 m3.
A motorboat traveling 5 m/s, East encounters a current traveling 2.5 m/s, South. What is the resultant velocity of the motor boat?b. If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?c. What distance downstream does the boat reach the opposite shore? The resultant velocity can be found using the Pythagorean theorem. The resultant is the hypotenuse of a right triangle with sides of 5 m/s and 2.5 m/s. It isSQRT (5 m/s) 2 + (2.5 m/s) 2 = 5.59 m/sIts direction can be determined using a trigonometric function.Direction = 360 degrees - invtan (2.5 m/s) / (5 m/s) = 333.4 degreesNOTE: the direction of the resultant velocity (like any vector) is expressed as the counterclockwise angle of rotation from due East.b.
The time to cross the river is t = d / v = (80 m) / (5 m/s) = 16.0 sc. The distance traveled downstream is d = v. t = (2.5 m/s). (16.0 s) = 40 m4.
A motorboat traveling 6 m/s, East encounters a current traveling 3.8 m/s, South. What is the resultant velocity of the motor boat?b. If the width of the river is 120 meters wide, then how much time does it take the boat to travel shore to shore?c. What distance downstream does the boat reach the opposite shore?
The resultant velocity can be found using the Pythagorean theorem. The resultant is the hypotenuse of a right triangle with sides of 6 m/s and 3.8 m/s. It isSQRT (6 m/s) 2 + (3.8 m/s) 2 = 7.10 m/sIts direction can be determined using a trigonometric function.Direction = 360 degrees - invtan (3.8 m/s) / (6 m/s) = 327.6 degreesNOTE: the direction of the resultant velocity (like any vector) is expressed as the counterclockwise direction of rotation from due East.b. The time to cross the river is t = d / v = (120 m) / (6 m/s) = 20.0 sc. The distance traveled downstream is d = v. t = (3.8 m/s). (20.0 s) = 76 m5.
If the current velocity in question #4 were increased to 5 m/s, then a. How much time would be required to cross the same 120-m wide river?b. What distance downstream would the boat travel during this time?