Physics vector

A motorboat heads due west at 10.0 m/s. Draw a scale vector diagram of the velocity vectors. The river has a current of 6.0 m/s due south. What is the resultant velocity of the boat?

11.7 m/s

What are the Pythagorean theorem

A^2 + B^2 = C^2

If the mass of the crate is 45 kg, what is the rate of acceleration of the crate?

a = Fnet / m a = 395N / 45 kg a = 8.78 m/s2

A girl is pulling her sister on a sled on level ground. The girl pulls with a force of 16 newtons on the rope which makes an angle of 40 o with the ground, and keeps the sled moving at constant speed. a. Draw a free body diagram showing the forces acting on the sled. b. Find the magnitude of the horizontal and vertical components of the girl's forces acting on the sled. c. What is the magnitude of the friction force acting on the sled?

Constant speed so . . . Ff = FgirlX = Fgirl Cosθ Ff = 16 N Cos 40̊ Ff = 12.3 N

A motorboat traveling 5 m/s, East encounters a current traveling 2.5 m/s, North. a. 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?

a. 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 is SQRT[(5 m/s)2 + (2.5 m/s)2] = 5.59 m/s Its direction can be determined using a trigonometric function. Direction = invtan [ (2.5 m/s) / (5 m/s)] = 26.6 degrees b. The time to cross the river is t = d / v = (80 m) / (5 m/s) = 16.0 s c. The distance traveled downstream is d = v • t = (2.5 m/s)*(16.0 s) = 40 m

In the final game of last year's regular season, South was playing New Greer Academy for the Conference Championship. In the last play of the game, star quarterback Avery took a snap from scrimmage and scooted backwards (northwards) 8.0 yards. He then ran sideways (westward) out of the pocket for 12.0 yards before finally throwing a 34.0 yard pass directly downfield (southward) to Kendall for the game-winning touchdown. Determine the magnitude and direction of the ball's displacement.

28.6 yd, 24.8° West of South or 245.2° CCW This motion involves three parts. They are shown in the graphic at the right. The resultant is the hypotenuse of a right triangle that has lengths of 26.0 yd, South (from 8.0 yd, North + 34.0 yd, South) and 12.0 yd, West. The Pythagorean theorem can be used to determine the magnitude of the resultant. The work is shown below. R2 = (26.0 yd)2 + (12.0 yd)2 = 820 yd 2 R = SQRT(820 yd2) R = 28.6 yd The tangent function can be used to determine the angle that the resultant makes with due south. tangent(Θ) = (12.0 m)/(26.0 m) = 0.46153 ... Θ = tan-1 (0.46153 ...) Θ = ~24.8°

Suppose that the plane in question 1 was flying with a velocity of 358 km/hr in a direction of 146 degrees (i.e., 34 degrees north of west). If the Canadian border is still located a distance of 1500 km north of Chicago, then how much time would it take to cross the border?

7.5 hr The same strategy used to answer #1 would have to be used in question #2. First find the northward component of the plane's velocity. vNorth = 358 • sine (146 degrees) = 200 km/hr Once determined, the d = v • t equation can be used to determine the time that it takes the plane to reach the Canadian border. dNorth = vNorth • t t = dNorth / vNorth = (1500 km) / (200 km/hr) = 7.5 hr

During her recent trip to the grocery store, Claire de Iles walked 28 m to the end of an aisle. She then made a right hand turn and walked 12 m down the end aisle. Finally, she made another right hand turn and walked 12 m in the opposite direction as her original direction. Determine the magnitude of Claire's resultant displacement. (The actual direction - east, west, north, south are not the focus.)

: 20 m In this problem, we do not know which way Claire originally walked. So let's suppose that we call it South. If she started south, then upon making a right-hand-turn, she is now heading west. And after making a second right-hand-turn, she is heading North. The diagram at the right depicts the physical situation. The resultant displacement is the hypotenuse of a right triangle that has sides 12 m, West and 16 m, South (from 28 m, South + 12 m, North). The Pythagorean theorem can be used to determine the magnitude of the resultant. The work is shown below. R2 = (12 m)2 + (16 m)2 = 400 m 2 R = SQRT(400 m2) R = 20 m

Cameron Per (his friends call him Cam) and Baxter Nature are on a hike. Starting from home base, they make the following movements. A: 2.65 km, 140° CCW B: 4.77 km, 252° CCW C: 3.18 km, 332° CCW Determine the magnitude and direction of their overall displacement.

A 2.65 km 140° CCW (2.65 km)•cos(40°) = 2.030... km, West (2.65 km)•sin(40°) = 1.703... km, North B 4.77 km 252° CCW (4.77 km)•sin(18°) = 1.474... km, West (4.77 km)•cos(18°) = 4.536... km, South C 3.18 km 332° CCW (3.18 km)•cos(28°) = 2.808... km, East (3.18 km)•sin(28°) = 1.493... km, South Sum of A + B + C 0.696 km, West 4.326 km, South

A boat heads straight across a river. The river flows north at a speed of 3 m/s. If the river current were greater, then the time required for the boat to reach the opposite shore would not change. a. True b. False

A - True The flow of the water down the river and parallel to its banks has no affect on the across-the-river motion of the boat. Perpendicular components of motion are independent of each other.

A boat begins at point A and heads straight across a 60-meter wide river with a speed of 4 m/s (relative to the water). The river water flows north at a speed of 3 m/s (relative to the shore). The boat reaches the opposite shore at point C. Which of the following would cause the boat to reach the opposite shore at a location SOUTH of C? a. The boat heads across the river at 5 m/s. b. The boat heads across the river at 3 m/s. c. The river flows north at 4 m/s. d. The river flows north at 2 m/s. e. Nonsense! None of these affect the location where the boat lands.

A and D The only two variables which would effect the northward motion of the boat would be the northward velocity of the river and the time required to cross the river. The northward distance is simply given by the equation dNorth = vNorth • t A decrease in the northward river velocity would decrease the distance and a decrease in time would decrease the distance. If the boat traveled faster across the river, the time would be decreased.

A person walks 40.0 meters east then 100.0 meters south. Draw a vector diagram to scale and find the total distance and the displacement of the person.

Dist = 140 m Disp = 108 m

Skye is trying to make her 70.0 kg Saint Bernard go out the back door but the dog refuses to walk. If the coefficient of sliding friction between the dog and the floor is 0.50, how hard must Skye push in order to move the dog with a constant speed?

Fa =Ff =μFn Fa = (0.50) (686N) Fn =Fg =mg Fn = (70kg) (9.8 m/s2) Fn = 686 N

Rather than taking the stairs, Pierre gets from the second floor of his house to the first floor by sliding down the banister that is inclined at an angle of 30.0 ° to the horizontal. If Pierre has a mass of 45 kg and the coefficient of sliding friction between Pierre and the banister is 0.20, what is the force of friction impeding Pierre's motion down the banister? If the banister is made steeper (inclined at a larger angle), will this have any effect on the force of friction? If so, what?

Ff = μFn Ff = (0.20) (382 N) Ff = 76.4 N Fn = Fgy = mg sin 60̊ Fn = (45kg) (9.8 m/s2) Fn = 382 N A steeper incline will result in a smaller Fgy, therefore a smaller Fn.

Billy is riding a go-cart coasting down a hill at constant speed. Billy and the cart combined have a mass of 150 kg. The hill is at a 40° angle to the horizontal ground. a. Are the forces acting on Billy and the cart balanced or unbalanced? b. Draw a free body diagram (FBD) showing all of the forces acting on Billy and the go-cart. c. Draw an x-y coordinate axis on your FBD and draw the x and y components of the Fg force. d. What is the magnitude of the weight force? e. Find the x and y components of the weight force. f. What is the magnitude of the friction force? g. What is the magnitude of the normal force?

Fg = mg Fg = (150kg) ( 9.8 m/s2) Fg = 1470 N Fgx = Fg Cos θ Fgx = (1470 N) (Cos 50̊ ) Fgx = 944.9 N Fgy = Fg Sin θ Fgy = (1470 N) (Sin 50̊ ) Fgy = 1126 N Ff = Fgx = 944.9 N FN = Fgy = 1126 N

A plane flies northwest out of O'Hare Airport in Chicago at a speed of 400 km/hr in a direction of 150 degrees (i.e., 30 degrees north of west). The Canadian border is located a distance of 1500 km due north of Chicago. The plane will cross into Canada after approximately ____ hours. a. 0.13 b. 0.23 c. 0.27 d. 3.75 e. 4.33 f. 6.49 g. 7.50 h. None of these are even close.

G The plane has both a northward and a westward motion. The northward motion towards the Canadian border is dependent upon the component of velocity in the northern direction. To solve this problem, the northern component of the plane's velocity must first be determined. vNorth = 400 • sine (150 degrees) = 200 km/hr Once determined, the d = v • t equation can be used to determine the time that it takes the plane to reach the Canadian border. dNorth = vNorth • t t = dNorth / vNorth = (1500 km) / (200 km/hr) = 7.5 hr

Rose is sledding down an ice-covered hill inclined at an angle of 15° with the horizontal. If Rose and the sled have a combined mass of 54.0 kg, what is the force pulling them down the hill?

The force pulling Rose down the hill is the component of gravity that is parallel to the hill so . . . Fgx = Fg Cos θ Fgx = m g Cos θ Fgx = (54.0 kg) (9.8 m/s2) Cos 15̊ Fgx = 511.2 N

Max plays middle linebacker for South's football team. During one play in last Friday night's game against New Greer Academy, he made the following movements after the ball was snapped on third down. First, he back-pedaled in the southern direction for 2.6 meters. He then shuffled to his left (west) for a distance of 2.2 meters. Finally, he made a half-turn and ran downfield a distance of 4.8 meters in a direction of 240° counter-clockwise from east (30° W of S) before finally knocking the wind out of New Greer's wide receiver. Determine the magnitude and direction of Max's overall displacement.

The triangle's perpendicular sides have lengths of 4.6 meters and 6.756 meters. The length of the horizontal side (4.6 m) was determined by adding the values of B (2.2 m) and Cx (2.4 m). The length of the vertical side (6.756... m) was determined by adding the values of A (2.6 m) and Cy (4.156... m). The resultant's magnitude (R) can now be determined using the Pythagorean theorem. R2 = (6.756... m)2 + (4.6 m)2 R2 = 45.655... m2 + 21.16 m2 R2 = 66.815... m2 R = SQRT(66.815... m2 ) R = 8.174 ... m R = ~8.2 m The direction of the resultant can be determined by finding the angle that the resultant makes with either the north-south or the east-west vector. The diagram at the right shows the angle theta (Θ) marked inside the vector addition triangle. This angle theta is the angle that the resultant makes with west. Its value can be determined using the tangent function. The tangent function (as in TOA) relates the angle value to the ratio of the lengths of the opposite side to the adjacent side. That is, tangent(Θ) = (6.756... m)/(4.6 m) = 1.46889... Using the inverse tangent function, the angle theta (Θ) can be determined. On most calculators, this involves using the 2nd-Tangent buttons. Θ = tan-1 (1.46889...) = 55.7536... ° Θ = ~56°

A 2750 kg rocket is lifting off and is accelerating at a rate of 35 m/s2. a. Draw a free body diagram showing the forces acting on the rocket during liftoff. b. What is the net force acting on the rocket? c. What is the magnitude of the thrust force exerted by the rocket's engine.

a = 35 m/s2 m = 2750 kg Fnet =m a Fnet = (2750 kg) (35 m/s2) Fnet = 96 250 kg m/s2 Fnet = 96 250 N Fnet = Fengine - Fg Fengine = Fg + Fnet Fg = m g = (2750 kg) (9.8 m/s2) Fg = 26 950 N Fengine = 26 950N + 96 250 N Fengine = 123 200N

If the mass of the crate is 45 kg, what is the rate of acceleration of the crate?

a = Fnet / m a = 60N / 45 kg a = 1.33 m/s2

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. 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 ] )

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?

a. It would require the same amount of time as before (20 s). Changing the current velocity does not affect the time required to cross the river since perpendicular components of motion are independent of each other. b. The distance traveled downstream is d = v • t = (5 m/s) • (20.0 s) = 100 m. Note that an alteration in the current velocity would only affect the distance traveled downstream (and the resultant velocity).

A motorboat traveling 4 m/s, East encounters a current traveling 7.0 m/s, North. a. What is the resultant velocity of the motorboat? 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?

a. 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 is SQRT [ (4 m/s)2 + (7 m/s)2 ] = 8.06 m/s Its 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 s c. The distance traveled downstream is d = v • t = (7 m/s) • (20 s) = 140 m

A motorboat traveling 5 m/s, East encounters a current traveling 2.5 m/s, South. a. 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?

a. 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 is SQRT [ (5 m/s)2 + (2.5 m/s)2 ] = 5.59 m/s Its direction can be determined using a trigonometric function. Direction = 360 degrees - invtan[ (2.5 m/s) / (5 m/s) ] = 333.4 degrees NOTE: 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 s c. The distance traveled downstream is d = v • t = (2.5 m/s) • (16.0 s) = 40 m

Question 4. A motorboat traveling 6 m/s, East encounters a current traveling 3.8 m/s, South. a. 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?

a. 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 is SQRT [ (6 m/s)2 + (3.8 m/s)2 ] = 7.10 m/s Its direction can be determined using a trigonometric function. Direction = 360 degrees - invtan[ (3.8 m/s) / (6 m/s) ] = 327.6 degrees NOTE: 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 s c. The distance traveled downstream is d = v • t = (3.8 m/s) • (20.0 s) = 76 m