I say that because it's the most likely thing that l wil pick
I think your talking about refraction and if so the first medium it existed in before entering the second usually has the higher speed especially it the first medium is less denser than the second
Answer:
a) 20 m/s
b) 37.5 m)s
Explanation:
Average speed = total distance ÷ total time
=> (a) average speed of a car that travels 400m in 20s
= 400/20 = 20 m/s
& (b) average speed of an athlete who runs 1500m in 4 minutes (or 4×60=240 seconds)
= 1500/240 = 37.5 m/s
Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
Answer:
<em>The distance the car traveled is 21.45 m</em>
Explanation:
<u>Motion With Constant Acceleration
</u>
It occurs when an object changes its velocity at the same rate thus the acceleration is constant.
The relation between the initial and final speeds is:
![v_f=v_o+at\qquad\qquad [1]](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat%5Cqquad%5Cqquad%20%5B1%5D)
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
The distance traveled by the object is given by:
![\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3Dv_o.t%2B%5Cfrac%7Ba.t%5E2%7D%7B2%7D%5Cqquad%5Cqquad%20%5B2%5D)
Solving [1] for a:
Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:
The distance is now calculated with [2]:
x = 21.45 m
The distance the car traveled is 21.45 m
