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:
C) 6 m/s
Explanation:
Given that
m₁=5000 kg
The initial velocity of 5000 kg car =u₁
m₂=10,000 kg
The initial velocity of 10000 kg car =u₂ = 0 m/s
After collision the final speed of the both car,v = 2 m/s
There is no any external force on the system that is why linear momentum will be conserved.
Linear momentum P = m v
m₁u₁ + m₂u₂ = (m₂ + m₁) v
5000 x u₁ + 10000 x 0 = (5000 + 10000) x 2
5000 x u₁ = 15000 x 2
5 x u₁ = 15 x 2
u₁ = 6 m/s
Therefore the answer is C.
C) 6 m/s
C. when you can't achieve your goal due to events beyond your control
The figure is showing a volume of 2.4 mL becuase it's feel 4 little segments.
Therefore, the answer is 2.4 mL.
