Answer:
The runner's speed at the following times would remain 8.64 m/s.
Explanation:
Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.
In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.
First equation of motion: Vf = Vi + a×t
Vf stands for final velocity
Vi stands for initial velocity
a stands for acceleration
t stands for time
In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.
The acceleration (a) is given as 2.4 m/s²
The time (t) is given as 3.6 s
Now put the values of Vi, a and t in first equation of motion
Vf = Vi + a×t
Vf = 0 + 2.4×3.6
Vf = 2.4×3.6
Vf = 8.64 m/s
So,the runner's speed at the following times would remain 8.64 m/s.
On the list of choices that you provided, there is no such statement.
a. 46 m/s east
The jet here is moving with a uniform accelerated motion, so we can use the following suvat equation to find its velocity:
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
where
v is the velocity calculated at time t
u is the initial velocity
a is the acceleration
The jet in the problem has, taking east as positive direction:
u = +16 m/s is the initial velocity
is the acceleration
Substituting t = 10 s, we find the final velocity of the jet:
And since the result is positive, the direction is east.
b. 310 m
The displacement of the jet can be found using another suvat equation
where
s is the displacement
u is the initial velocity
a is the acceleration
t is the time
For the jet in this problem,
u = +16 m/s is the initial velocity
is the acceleration
t = 10 s is the time
Substituting into the equation,
![s=(16)(10)+\frac{1}{2}(3)(10)^2=310 m](https://tex.z-dn.net/?f=s%3D%2816%29%2810%29%2B%5Cfrac%7B1%7D%7B2%7D%283%29%2810%29%5E2%3D310%20m)
The charge of the copper nucleus is 29 times the charge of one proton:
![Q=29 q= 29 \cdot 1.6 \cdot 10^{-19}C=4.64 \cdot 10^{-18}C](https://tex.z-dn.net/?f=Q%3D29%20q%3D%2029%20%5Ccdot%201.6%20%5Ccdot%2010%5E%7B-19%7DC%3D4.64%20%5Ccdot%2010%5E%7B-18%7DC)
the charge of the electron is
![e=-1.6 \cdot 10^{-19}C](https://tex.z-dn.net/?f=e%3D-1.6%20%5Ccdot%2010%5E%7B-19%7DC)
and their separation is
![r=1.0 \cdot 10^{-12} m](https://tex.z-dn.net/?f=r%3D1.0%20%5Ccdot%2010%5E%7B-12%7D%20m)
The magnitude of the electrostatic force between them is given by:
![F=k_e \frac{Qe}{r^2}](https://tex.z-dn.net/?f=F%3Dk_e%20%20%5Cfrac%7BQe%7D%7Br%5E2%7D%20)
where
![k_e](https://tex.z-dn.net/?f=k_e)
is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)