Calculate the acceleration of the car which moves 36m/s in 8s
1 answer:
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The formula for velocity vf = vi + at
First list your given information
2m/s Is your initial velocity (vi)
6m/s is you final velocity (vf)
2 seconds is your time (t)
Since you want the a for acceleration get a by itself
a = (vf-vi)/t
So a= (6-2)/2
a= 4/2
a=2
Now units
the units for acceleration are m/s
2m/s
First list your given information
2m/s Is your initial velocity (vi)
6m/s is you final velocity (vf)
2 seconds is your time (t)
Since you want the a for acceleration get a by itself
a = (vf-vi)/t
So a= (6-2)/2
a= 4/2
a=2
Now units
the units for acceleration are m/s
2m/s
Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.
Explanation:
Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.
Now, according to the law of conservation of energy the formula used is as follows.
As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.
Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.