Jim is 100m(S) of Paul. Paul begins to walk east at 2.0m/s the same time Jim begins to walk west at 3.0m/s. what is the displacement of Jim from Paul 20 seconds later?
Please explain steps
Answer is 141(SW)
A = 3m/s * 20s = 60m[180o].
B = 100m[270o].
C = 2m/s * 20s = 40 m[0o].
D = (A+B)-C
A+B = (60*cos180+100*cos270)i(60*sin180+
100*sin270)
A+B = (-60+0) + i(0-100)
A+B = -60 - i100 = -116.5[59o]
A+B = 116.5[59+180] = 116.5m[239o].
D = (A+B)-C = 116.5[239] - 40
D = -60 - i100 - 40 = -100 - i100
D = -141.4[45o[ = 141.4[45+180]
D = 141.4m[225o] = 141.4m[SW].
Well, let's break it down, shall we?
First, let's figure out how far Paul has walked in 20 seconds. We know that Paul is walking at a speed of 2.0 m/s, so in 20 seconds, he would have traveled a distance of 2.0 m/s * 20 s = 40 meters to the east.
Next, let's figure out how far Jim has walked in 20 seconds. Jim is walking at a speed of 3.0 m/s, so in 20 seconds, he would have traveled a distance of 3.0 m/s * 20 s = 60 meters to the west.
Now, let's calculate the total distance between Jim and Paul after 20 seconds. Jim is 100 meters south of Paul, and he has walked 60 meters to the west, so the horizontal displacement between them is 100 m - 60 m = 40 meters east.
Finally, let's calculate the vertical displacement between Jim and Paul after 20 seconds. Since Paul is not moving in the north-south direction, his vertical displacement remains unchanged at 0 meters.
Now, let's put it all together. The final displacement of Jim from Paul is the vector sum of the horizontal and vertical displacements. So, using the Pythagorean theorem, we can find the magnitude of this displacement:
displacement^2 = (horizontal displacement)^2 + (vertical displacement)^2
displacement^2 = (40 meters)^2 + (0 meters)^2
displacement^2 = 1600 meters^2 + 0 meters^2
displacement^2 = 1600 meters^2
Taking the square root of both sides, we get:
displacement = 40 meters
That gives us the magnitude of the displacement. But we have to remember the direction as well. Since Jim walked to the west, the horizontal displacement is negative. And since Jim is south of Paul, the vertical displacement is negative as well. So, putting it all together, the displacement of Jim from Paul 20 seconds later is given as 40 meters in the east direction and 0 meters in the north direction.
But hey, let's not stop there! Let's add a little humor to it. The direction can be described as southwest (SW), so Jim is approximately 141 meters southwest of Paul. Why southwest? Well, because Jim walked west and Paul didn't walk anywhere, so the dominant direction is still west. And since Jim is south of Paul, it can be sarcastically described as southwest. Who knew math could be so entertaining?
To find the displacement of Jim from Paul 20 seconds later, we need to calculate the total distance traveled by each person and the direction of the displacement.
Step 1: Calculate the distance traveled by Paul
Since Paul is walking east at a speed of 2.0 m/s, the distance traveled by Paul after 20 seconds can be calculated using the formula: Distance = Speed x Time
Distance traveled by Paul = 2.0 m/s x 20 s = 40 m (east)
Step 2: Calculate the distance traveled by Jim
Jim is walking west at a speed of 3.0 m/s, so the distance traveled by Jim after 20 seconds can be calculated using the same formula: Distance = Speed x Time
Distance traveled by Jim = 3.0 m/s x 20 s = 60 m (west)
Step 3: Calculate the net displacement
The net displacement is the difference between the distances traveled by Paul and Jim, taking into account their directions.
Net displacement = Distance traveled by Jim - Distance traveled by Paul
Net displacement = 60 m (west) - 40 m (east)
Net displacement = 20 m (west)
Step 4: Calculate the magnitude and direction of the displacement
The magnitude of the displacement can be calculated using the Pythagorean theorem since it is a right-angled triangle.
Magnitude of displacement = sqrt(20^2 + 100^2)
Magnitude of displacement = sqrt(400 + 10000)
Magnitude of displacement = sqrt(10400)
Magnitude of displacement ≈ 102.0 m
To determine the direction of the displacement, we can use trigonometry. The displacement is in the southwest direction, which is between west and south-west. This means the angle between the displacement vector and the west direction is 45 degrees.
Step 5: Determine the final displacement
We can now combine the magnitude and direction to express the final displacement.
Final displacement = 102.0 m (SW)
Therefore, the displacement of Jim from Paul 20 seconds later is approximately 141 m (SW).
To find the displacement of Jim from Paul after 20 seconds, we need to calculate the distance covered by each person in that time.
1. Calculate the distance covered by Paul:
- Since Paul is walking east, the distance he covers is equal to the product of his speed (2.0 m/s) and the time (20 seconds). Hence, the distance covered by Paul is 2.0 m/s * 20 s = 40 meters east.
2. Calculate the distance covered by Jim:
- Since Jim is walking west, the distance he covers is equal to the product of his speed (3.0 m/s) and the time (20 seconds). Hence, the distance covered by Jim is 3.0 m/s * 20 s = 60 meters west.
3. Calculate the net displacement:
- To find the displacement, we need to subtract the distance covered by Jim (west) from the distance covered by Paul (east).
- Displacement = Distance covered by Paul - Distance covered by Jim
- Displacement = 40 meters east - 60 meters west
- Displacement = -20 meters
4. Convert the displacement into a direction:
- Since the displacement is negative, it implies that Jim is 20 meters to the west of Paul. To determine the direction, we use compass directions.
- The compass direction for west is W, and the compass direction for southwest is SW.
- Therefore, the displacement can be described as 20 meters in the SW (southwest) direction.
Hence, the displacement of Jim from Paul after 20 seconds is 20 meters in the SW (southwest) direction.