Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
![\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bate%5E%7B-6t%7D%5D%3Da%5B%281%29e%5E%7B-6t%7D%2Bt%28e%5E%7B-6t%7D%28-6%29%29%5D)
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
![a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}](https://tex.z-dn.net/?f=a%5B%281%29e%5E%7B-6t%7D-6te%5E%7B-6t%7D%5D%3D0%5C%5C%5C%5C1-6t%3D0%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B6%7D)
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
hence, the maximum speed is v_max = ((1/6)e^-1)a
Field lines become straight and perpendicular because every point of circular loop the concentric circles become larger and larger as we move away from the wire.
The answer should be 1.6 kilometers.
Answer:
None, both objects will hit ground at the same time.
Explanation:
- Assuming no air resistance present, and that both objects start from rest, we can apply the following kinematic equation for the vertical displacement:
- As the left side in (1) is the same for both objects, the right side will be the same also.
- Since g is constant close to the surface of the Earth, it's also the same for both objects.
- So, the time t must be the same for both objects also.
Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration:
, and where the negative sign means it points downwards, against the direction of the motion.
Therefore, we can use the following formula to solve the problem:
where
is the initial vertical velocity of the athlete,
is the vertical velocity of the athlete at the maximum height (and
at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
