Crucial Substitution

Classical Mechanics Level 3

The motor Of an Electric Train can give it an acceleration of 1 m/s 2 1 \text{ m/s}^{2} and It's brakes Can give it a negative acceleration of 3 m/s 2 3 \text{ m/s}^{2} . What is the shortest time in which the train can make a trip between two stations 1215 m 1215 \text { m} apart is?


The answer is 56.9.

Mehul Arora by Mehul Arora , 20, India

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3 solutions

Since the highest velocity reached from 0 by time T a is the same as the velocity destroyed by retardation in T r . 1 ∗ T a = 3 ∗ T r . ⟹ T r = T a 3 . T o t a l t i m e T = T a + T r = 4 3 ∗ T a . ∴ 1215 = . 5 ∗ 1 ∗ T a 2 + . 5 ∗ 3 ∗ ( T a 3 ) 2 . . . . . . S o l v i n g T = 56.92 \text{Since the highest velocity reached from 0 by time } T_a \text{ is the same}\\\text{as the velocity destroyed by retardation in } T_r.\\1*T_a=3*T_r.~\implies~T_r=\dfrac {T_a} 3.~Total ~time~T=T_a+T_r=\dfrac{ 4}{ 3} *T_a.\\\therefore~1215=.5*1*T_a^2+.5*3*(\dfrac{T_a} {3} )^2......Solving~~T=~~~~~ \color{#D61F06}{ \Large 56.92}

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Exact calculation is 3224 \sqrt{3224} , nearly 56.78.

Rajen Kapur - 6 years ago

please make me understand from calculating total time .......how did you find Ta+Tr =4/3 * Ta???

Oliul Taj - 6 years ago

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Highest velocity is same for acceleration and retardation say V with respective times as
T a a n d T r . V e l o c i t y = a c c . ( o r r e t . ) × t i m e . ∴ 1 × T a = 3 × T r . ∴ T r = T a 3 . B u t T o t a l t i m e = T a + T r = 4 3 ∗ T a . T_a~ and~T_r. Velocity=acc.(or~~ ret.) \times ~time.\\ \therefore~1\times T_a =3 \times T_r.\\\therefore~T_r=\dfrac {T_a} 3.~~But~Total~time=T_a+T_r=\dfrac{ 4}{ 3} *T_a.

Niranjan Khanderia - 6 years ago

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thnx for telling me

Oliul Taj - 6 years ago

Hey ..excuse me but u r requested to check your question further before giving a suitable solution.... As I am getting a total of 39.2secs

Rahul Bhattacharya - 5 years, 12 months ago

What is the relative velocity of a whale with respect to a ship floating on water at speed √x m/s????

Rahul Bhattacharya - 5 years, 12 months ago
Tijmen Veltman
May 4, 2015

The average speed is highest (and hence the time shortest) if the train accelerates for as long as possible and then brakes to come to a complete stop just in time. If v v is the maximal speed in this case, the distance traveled is equal to 1 2 v 2 (while accelerating) + 1 6 v 2 (while decelerating) = 2 3 v 2 = 1215 m \frac12 v^2 \text{(while accelerating)}+\frac16 v^2 \text{(while decelerating)}=\frac23 v^2=1215 \text{m} . Hence v = 3 2 ⋅ 1215 ≈ 42.7 m/s v=\sqrt{\frac32 \cdot 1215}\approx 42.7 \text{m/s} . This gives us an average speed of v / 2 = 21.3 m/s v/2=21.3 \text{m/s} and a time of 1215 m 21.3 m/s ≈ 56.9 s \frac{1215\text{m}}{21.3 \text{m/s}}\approx\boxed{56.9\text{s}} .

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Excellent! @Tijmen Veltman

Swapnil Das - 6 years ago
Lu Chee Ket
Oct 25, 2015

Triangle rather than trapezium on the graph of v versus t with highest speed for shortest time.

1 m/ s per 1 s means 3 T m/ s per 3 T s with acceleration of 1 m/ s^2 on climbing side;

There shall be -3 T m/ s per T s with acceleration of - 3 m/ s^2 on the remained side.

Area under the curve such as (1/ 2)(3 T + T)(3 T) = 1215;

T = Sqrt (405/ 2) => Shortest time = 4 T = 18 Sqrt (10) = 56.920997883030827975980083799789 s

Crucial substitution means trapezium is substituted with triangle.

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