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jeka94
3 years ago
5

A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of

85 km/h, in seconds, starting from rest? Hints deduction per hint. Hints remaining 1 Feedback: deduction per feedback A 33 Part (b) The same train ordinarily brakes at a rate of 1.8 m/s. How long does it take to come to a stop, in seconds, from its top speed? 433 Part (c) In emergencies the train can slow more rapidly. coming to rest from 85 km/h in 7.9 5. What is the magnitude of the acceleration during emergency braking, in meters per square second?
Physics
1 answer:
Dennis_Churaev [7]3 years ago
6 0

Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = 85\frac{1000}{3600}=23.61\ m/s

Initial velocity = u = 0

Equation of motion

v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2

Emergency acceleration is 2.99 m/s² (magnitude)

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3 years ago
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\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here's the solution ~

As we know, Displacement =

\qquad \sf  \dashrightarrow \: velocity  \times time

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<h3>OR </h3>

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8 0
2 years ago
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Answer: Hello the missing piece of your question is attached

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