To start the control would be import to see how drivers would act with full focus on the road. The subjects should (in order for the only difference to be the phone) is to use the same vehicle same road same weather condition, just obviously one has the phone the other does not. I hope this helps
Hi there!
We can use the following (derived) equation to solve for the final velocity given height:
vf = √2gh
We can rearrange to solve for height:
vf² = 2gh
vf²/2g = h
Plug in the given values (g = 9.81 m/s²)
(13)²/2(9.81) = 8.614 m
We can calculate time using the equation:
vf = vi + at, where:
vi = initial velocity (since dropped from rest, = 0 m/s)
a = acceleration (in this instance, due to gravity)
Plug in values:
13 = at
13/a = t
13/9.81 = 1.325 sec
If the other driver is going speed limit you can't pass him but if he's going slower than the limit you have to go in the left lane i'm not sure by how much i'll guess 5mph<span />
Explanation:
a) Power = work / time = force × distance / time
P = Fd/t
P = (85 kg × 9.8 m/s²) (4.6 m) / (12 s)
P ≈ 319 W
b) P = Fd/t
0.70 (319 W) = (m × 9.8 m/s²) (4.6 m) / (9.6 s)
m = 47.6 kg
Answer:
Av = 25 [m/s]
Explanation:
To solve this problem we must use the definition of speed, which is defined as the relationship between distance over time. for this case we have.
where:
Av = speed [km/h] or [m/s]
distance = 180 [km]
time = 2 [hr]
Therefore the speed is equal to:
![Av = \frac{180}{2} \\Av = 90 [km/h]](https://tex.z-dn.net/?f=Av%20%3D%20%5Cfrac%7B180%7D%7B2%7D%20%5C%5CAv%20%3D%2090%20%5Bkm%2Fh%5D)
Now we must convert from kilometers per hour to meters per second
![90[\frac{km}{h}]*1000[\frac{m}{1km}]*1[\frac{h}{3600s} ]= 25 [m/s]](https://tex.z-dn.net/?f=90%5B%5Cfrac%7Bkm%7D%7Bh%7D%5D%2A1000%5B%5Cfrac%7Bm%7D%7B1km%7D%5D%2A1%5B%5Cfrac%7Bh%7D%7B3600s%7D%20%5D%3D%2025%20%5Bm%2Fs%5D)
