1.53 m/s towards the beach
Magnitude of average acceleration = (change of speed) divided by (time for the change)Average 'A' = (6 - 4) / 20 = 2/20 = 0.1 meter per second2-- That's the average over the 20 seconds. We don't know anything about thevalue of the acceleration at any particular instant during the 20 seconds.-- We're working entirely with scalars ... speed, not velocity, and magnitude ofacceleration ... since we don't know anything about the runner's direction atany time during the whole event.
6 m/s divided by 12 s or 0.5 m/s^2. (a half meter per second per second) (on average)
Average acceleration over a certain time period is the difference in velocity, divided by the time. In this case, (4-2)/W.
Assuming that acceleration is constant during that time, just divide the change in speed by the time.
If a runner runs a mile in 4 minutes 30 seconds he is averaging 13.33 MPH over the total distance.
1.53 m/s towards the beach
Magnitude of average acceleration = (change of speed) divided by (time for the change)Average 'A' = (6 - 4) / 20 = 2/20 = 0.1 meter per second2-- That's the average over the 20 seconds. We don't know anything about thevalue of the acceleration at any particular instant during the 20 seconds.-- We're working entirely with scalars ... speed, not velocity, and magnitude ofacceleration ... since we don't know anything about the runner's direction atany time during the whole event.
Its usually takes an athlete (preferably Usain bolt) 29.9 Seconds to run 300 meters so a high school runner around 39 secs and a college student around 34 seconds depends on the age to be honset and stamana but roughly 30 to 40 seconds 1 minute for the max .
It is 0.1 metre per second-squared.
About 365.5 meters a minute.
It is 0.1 metre per second-squared.
It is 0.1 metre per second-squared.
It is 0.1 metre per second-squared.
6 m/s divided by 12 s or 0.5 m/s^2. (a half meter per second per second) (on average)
40 metres in 8seconds what would the speed be
3.9 nm = how many meters
50