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Answer:
The object will travel 675 m during that time.
Explanation:
A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.
In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.
In this case, the position is calculated using the expression:
x = xo + vo*t + ½*a*t²
where:
- x0 is the initial position.
- v0 is the initial velocity.
- a is the acceleration.
- t is the time interval in which the motion is studied.
In this case:
- x0= 0
- v0= 0 because the object is initially stationary
- a= 6
- t= 15 s
Replacing:
x= 0 + 0*15 s + ½*6 *(15s)²
Solving:
x=½*6 *(15s)²
x=½*6 *225 s²
x= 675 m
<u><em> The object will travel 675 m during that time.</em></u>
Answer:
6.5 m/s
Explanation:
We are given that
Distance, s=100 m
Initial speed, u=1.4 m/s
Acceleration,
We have to find the final velocity at the end of the 100.0 m.
We know that
Using the formula
Hence, her final velocity at the end of the 100.0 m=6.5 m/s
Answer:
6370 J
Explanation:
By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m
where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference