A man completes 31 lengths of a 50 metre pool with an average speed of 0.9 km/h. What is the magnitude of his average velocity vector in Standard SI units?
Velovicy is a VECTOR
north one way, south back
31 lengths, 30 of which were back and forth
so
went 50 meters total north in the end
How long did it take?
31 * 50 = 1550 meters back and forth at 900 m/3600 s
time = 1550 *3600/900
= 6200 seconds
so 50 meters/6200 seconds
= 0.00806 m/s NORTH
To find the magnitude of the average velocity vector, we need to convert the average speed from km/h to m/s.
Given that the average speed is 0.9 km/h, we can convert it to m/s by multiplying it by 1000/3600, since there are 1000 meters in a kilometer and 3600 seconds in an hour.
So, average speed = 0.9 km/h * (1000 m / 3600 s)
= 250/900 m/s
= 5/18 m/s
To find the magnitude of the average velocity vector, we need to multiply the average speed by the number of lengths completed. In this case, the man completed 31 lengths, each of which is 50 meters long.
Magnitude of average velocity vector = (5/18 m/s) * (31 * 50 m)
= 775/18 m/s
≈ 42.43 m/s
Therefore, the magnitude of the man's average velocity vector in standard SI units is approximately 42.43 m/s.
To find the magnitude of the average velocity vector, we need to determine the total displacement and divide it by the total time taken.
First, let's calculate the total distance covered by the man. Since he completes 31 lengths of a 50-meter pool, his total distance covered is:
Distance = (31 lengths) x (50 meters/length) = 1550 meters
Next, we need to convert the average speed from km/h to m/s, which are the standard SI units for velocity. To do this, we can use the conversion factor:
1 km/h = (1 km/h) x (1000 m/km) / (3600 s/h) = 0.2778 m/s
So, the man's average speed in m/s is:
Speed = (0.9 km/h) x (0.2778 m/s) = 0.25002 m/s (rounded to five decimal places)
Now, we can find the total time taken. To do this, we divide the total distance by the average speed:
Time = Distance / Speed = 1550 meters / 0.25002 m/s = 6199.356 s (rounded to three decimal places)
Finally, we can find the magnitude of the average velocity vector by dividing the total displacement by the total time taken:
Magnitude of Average Velocity = Distance / Time = 1550 meters / 6199.356 s ≈ 0.25002 m/s
Therefore, the magnitude of the man's average velocity vector in standard SI units is approximately 0.25002 m/s.