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>
<span>When an object moves in a circle, the acceleration points toward the center of the circle. This acceleration is called centripetal acceleration.
We can use a simple equation to find centripetal acceleration.
a = v^2 / r
We can use this same equation to find the speed of the car.
v^2 = a * r
v = sqrt { a * r }
v = sqrt{ (1.50)(9.80 m/s^2)(11.0 m) }
v = 12.7 m/s
The speed of the roller coaster is 12.7 m/s</span>
Answer:
v₂=- 34 .85 m/s
v₁=0.14 m/s
Explanation:
Given that
m₁=70 kg ,u₁=0 m/s
m₂=0.15 kg ,u₂=35 m/s
Given that collision is elastic .We know that for elastic collision
Lets take their final speed is v₁ and v₂
From momentum conservation
m₁u₁+m₂u₂=m₁v₁+m₂v₂
70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂
70 x v₁ + 0.15 x v₂=5.25 --------1
v₂-v₁=u₁-u₂ ( e= 1)
v₂-v₁ = -35 --------2
By solving above equations
v₂=- 34 .85 m/s
v₁=0.14 m/s
The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².
<h3>What will be the maximum acceleration of the truck to avoid tipping over?</h3>
The maximum acceleration is obtained by taking clockwise moments about the tipping point of rotation.
Clockwise moment = Anticlockwise moment
Ft * 1.58 m = F * 0.67 m
where
- Ft is tipping force = mass * acceleration, a
- F is weight = mass * acceleration due to gravity, g
m * a * 1.58 = m * 9.81 * 0.67
a = 4.15 m/s²
The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².
In conclusion, the acceleration of the truck is found by taking moments about the tipping point.
Learn more about moments of forces at: brainly.com/question/27282169
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Answer:
d
Explanation:
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