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>
m = 5 kg
a = 2 m/s²
to find the force that accelerates the 4 kg object @ 2 m/s²
F = ma = 5 kg x 2 m/s² = 10 N
To find what acceleration 10 N would give a 20 kg object
a = F/m = 10 N/20 kg = 0.5 m/s
Answer:
0.66
Explanation:
By using the formula
u = v^2 / r g
Where u is coefficent of friction
u = 23.5 × 23.5 / (85 × 9.8)
u = 0.66
<span> the </span>electric field<span> direction about a </span>positive<span> source </span>charge<span> is always directed away from the </span>positive<span> source. And the </span>electric field <span>direction about a negative source </span>charge<span> is always directed toward the negative source.</span>
