1) The velocity of the particle is given by the derivative of the position. So, if we derive s(t), we get the velocity of the particle as a function of the time:
2) The acceleration of the particle is given by the derivative of the velocity. So, if we derive v(t), we get the acceleration of the particle as a function of the time:
Answer:
U = 30 m/s
a = 3 m/s²
Explanation:
"A car accelerating uniformly from rest reaches a maximum speed of U in 10 s. It then moves with that speed for an additional 20 s. The distance covered by the car in the 30 s interval is 750 m. Find U and the acceleration of the car in the first 10 s."
During the first 10 s:
v₀ = 0 m/s
v = U m/s
t = 10 s
The distance covered in this time is the average velocity times time:
Δx = ½ (v + v₀) t
Δx = ½ (U + 0) (10)
Δx = 5U
The distance covered in the next 20 seconds is speed times time:
Δx = 20U
The total distance is 750 m:
5U + 20U = 750
25U = 750
U = 30 m/s
The acceleration during the first 10 seconds is the change in speed over change in time.
a = Δv / Δt
a = (30 m/s − 0 m/s) / 10 s
a = 3 m/s²
I hope this helps. Your answer is D
Hi!
The answer would be 22.5m
<h3>Explanation</h3>
To calculate the distance the ball travels, you would need to apply the following formula from the equations of motion for an accelerating object:
2as = v^2 - u^2
Where a is acceleration due to gravity, 9.8m/s^2
s is the distance traveled by the object (height of ball in this case)
v is the final velocity, which we know will be zero at the point where the ball reaches maximum height.
u is the initial velocity, which is known to us as 21m/s
Rearranging the equation to solve for the height:
s = (v^2 - u^2 ) / 2a
s = ( 0^2 - 21^2 ) / 2(-9.8)
s = - 441/ - 19.6
s = 22.5m
<em>Note: since gravity is acting against the object's motion, it will be negative </em>
<em>Hope this helps!</em>
Answer:
A. The sound wave will reflect off Buildings and automobiles.
Explanation:
This is because the sound waves would more likely propagate through diffraction through buildings and transmission through the air. It is also more likely to be absorbed by buildings than for multiple reflections to occur off buildings and automobiles. In the process of reflection, these materials would absorb the sound energy thereby reducing its ability to reflect.
