Answer: 10 s, 30 m/s , 150 m
Explanation:
Given
The speed of motorcyclist is ![u_1=15\ m/s](https://tex.z-dn.net/?f=u_1%3D15%5C%20m%2Fs)
The initial speed of a police motorcycle is ![u_2=0\ m/s](https://tex.z-dn.net/?f=u_2%3D0%5C%20m%2Fs)
acceleration of police motorcycle is ![a=3\ m/s^2](https://tex.z-dn.net/?f=a%3D3%5C%20m%2Fs%5E2)
Police will catch the motorcyclist when they traveled equal distances
distance traveled by motorcyclist in time t is
![\Rightarrow s_1=15\times t](https://tex.z-dn.net/?f=%5CRightarrow%20s_1%3D15%5Ctimes%20t)
Distance traveled by Police in time t is
![\Rightarrow s_2=u_2t+\frac{1}{2}at^2\\\Rightarrow s_2=0+0.5\times 3\times t^2](https://tex.z-dn.net/?f=%5CRightarrow%20s_2%3Du_2t%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%20s_2%3D0%2B0.5%5Ctimes%203%5Ctimes%20t%5E2)
put ![s_1=s_2](https://tex.z-dn.net/?f=s_1%3Ds_2)
![\Rightarrow 15t=0.5\times 3\times t^2\\\\\Rightarrow t(1.5t-15)=0\\\\\Rightarrow t=\dfrac{15}{1.5}=10\ s](https://tex.z-dn.net/?f=%5CRightarrow%2015t%3D0.5%5Ctimes%203%5Ctimes%20t%5E2%5C%5C%5C%5C%5CRightarrow%20t%281.5t-15%29%3D0%5C%5C%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B15%7D%7B1.5%7D%3D10%5C%20s)
Police officer's speed at that time is
![\Rightarrow v_2=u_2+at\\\Rightarrow v_2=0+3\times 10=30\ m/s](https://tex.z-dn.net/?f=%5CRightarrow%20v_2%3Du_2%2Bat%5C%5C%5CRightarrow%20v_2%3D0%2B3%5Ctimes%2010%3D30%5C%20m%2Fs)
Distance traveled by each vehicle is
![\Rightarrow s_1=s_2=15\times t=15\times 10=150\ m](https://tex.z-dn.net/?f=%5CRightarrow%20s_1%3Ds_2%3D15%5Ctimes%20t%3D15%5Ctimes%2010%3D150%5C%20m)
<span>d.rotating counterclockwise and slowing down
This is a matter of understanding the notation and conventions of angular rotations. Positive rotations are counter clockwise and negative rotations are clockwise. An easy way to remember this is the "right hand rule". Make a closed fist with your right hand and have the thumb sticking outwards. If you orient your thumb such that it's pointing in the direction of the positive value along the axis, your fingers will be curled in the positive rotational direction. So in the described scenario, the sphere is rotating in the positive direction (counter clockwise) and decelerating due to the negative angular acceleration. That immediately indicates that options "a", "b", and "e" are wrong since they mention the sphere going clockwise at the beginning. Of the two remaining options "c" and "d", we can discard option "c" since it has the rotation speeding up, and that leaves us with option "d" where the sphere is rotating counter clockwise and slowing down.</span>
The answer would be B.
<span>
Standard deviation basically measures how spread out the values are. Without solving, you can easily tell which one among your choices have a smaller deviation. The closer the values are to each other the smaller the standard deviation. The values of choice B are the closest together, so you can assume that they have the smallest standard deviation. </span>
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ =
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m