A barge is being towed south at the rate 2 km/hr. A man on the deck walks from west to east at the rate 4ft/s. Find the magnitude and direction of the man's actual velocity.
1 km = appr 3280.84 feet
2 km/h = 2(3280.84/3600) ft/s
= 1.82269 ft/s
let r be the resultant velocity
r^2 = 2^2 + 1.82269^2
r = √7.32219..
= appr 2.7 ft/s
tanØ = 1.82269/2 , where Ø is in my right-angled triangle
Ø = appr 42.34°
his direction is S 42.34° E
or by vectors
vector r = 2(cos270 , sin270) + 1.82269(cos0 , sin0)
= 2(0, -1) + 1.82269(1,0)
= (0,-2) + (1.82269,0)
= (1.82269,-2)
magnitude = √(1.82269^2 + (-2)^2)
= 2.7 as above
direction:
tan k = -2/1.82269
k = 312.34° which is the same as S 42.34 W
X = 4 Ft/s
Y = -2km/h = -2000m/3600s * 3.3Ft/m = -1.833 Ft/s
Tan A = Y/X = -1.833/4 = -0.45825
A = -24.62o = 24.62o South of East. =
65.38o East of South = Direction.
Magnitude = X/Cos A = 4/Cos(-24.62) =
4.4 Ft/s.
To find the magnitude and direction of the man's actual velocity, we need to combine the velocities of the barge and the man.
First, let's convert the units to be consistent. Since the barge's velocity is given in km/hr, we need to convert the man's velocity to km/hr as well.
1 foot = 0.0003048 kilometers
1 second = 1/3600 hours
The man's velocity is 4 ft/s, so to convert to km/hr, we multiply by the conversion factors:
4 ft/s * 0.0003048 km/ft * (3600 s/1 hr) = 0.4384 km/hr
Now that we have both velocities in km/hr, we can combine them.
The barge is moving south at a rate of 2 km/hr. We can represent the barge's velocity as a vector: V_barge = 2 km/hr * (0, -1) (since it's moving south).
The man is walking from west to east at a rate of 0.4384 km/hr. We can represent the man's velocity as a vector: V_man = 0.4384 km/hr * (1, 0) (since he's moving horizontally from west to east).
To find the man's actual velocity, we add these two vectors:
V_actual = V_barge + V_man
= (0, -1) + (1, 0)
= (1, -1)
So the man's actual velocity has a magnitude of 1 km/hr and a direction going northeast.
Therefore, the magnitude of the man's actual velocity is 1 km/hr and the direction is northeast.
To find the magnitude and direction of the man's actual velocity, we need to use vector addition.
The velocity of the barge being towed south is given as 2 km/hr. Since this velocity is in kilometers per hour, we need to convert the units of the man's velocity to kilometers per hour as well.
Given that the man walks at a rate of 4 ft/s, we need to convert this to km/hr.
First, let's convert 4 ft/s to km/hr:
1 mile = 5280 ft
1 hour = 3600 s
To convert feet to kilometers and seconds to hours, we can use the conversion factors:
1 km = 0.0003048 miles
1 hour = 3600 s
So, we have:
(4 ft/s) * (1 mile/5280 ft) * (0.0003048 km/1 mile) * (3600 s/1 hour) = 0.01344 km/hr
Therefore, the magnitude of the man's velocity is 0.01344 km/hr.
Now, let's find the direction of the man's actual velocity.
Since the barge is being towed south and the man is walking from west to east, the man's actual velocity will combine the velocity due to walking eastward with the velocity due to the barge towing southward.
To determine the resultant velocity, we can use the concept of vector addition. To do this, we can draw a vector diagram.
Let's denote the direction of the barge's velocity as south, and the direction of the man's velocity as east.
Using the Pythagorean theorem, we can find the resultant vector of the man's actual velocity, which is the hypotenuse of a right triangle formed by the barge's velocity vector and the man's velocity vector.
Given that the magnitudes of the man's velocity and the barge's velocity are 0.01344 km/hr and 2 km/hr, respectively, we can calculate the resultant velocity using the equation:
(resultant velocity)^2 = (man's velocity)^2 + (barge's velocity)^2
(resultant velocity)^2 = (0.01344 km/hr)^2 + (2 km/hr)^2
(resultant velocity)^2 = 0.0001804096 km^2/hr^2 + 4 km^2/hr^2
(resultant velocity)^2 = 4.0001804096 km^2/hr^2
Simplifying, we find:
(resultant velocity)^2 = 4.0001804096 km^2/hr^2
Taking the square root of both sides:
resultant velocity = sqrt(4.0001804096 km^2/hr^2)
resultant velocity ≈ 2.0000451 km/hr
Therefore, the magnitude of the man's actual velocity is approximately 2.0000451 km/hr.
To find the direction, we refer back to the vector diagram. The direction of the resultant vector can be found by measuring the angle it makes with the south direction.
Using trigonometry, we can calculate the angle:
tan(angle) = (man's velocity) / (barge's velocity)
tan(angle) = (0.01344 km/hr) / (2 km/hr)
angle = arctan(0.01344/2)
Using a calculator, we find:
angle ≈ 0.388473 degrees
Therefore, the direction of the man's actual velocity is approximately 0.388473 degrees east of south.