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    a train starting from a railway station and moving with uniform acceleration attains a speed of 40 km per hour in 10 minutes find its acceleration

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    A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km hr^

    Click hereπŸ‘†to get an answer to your question ✍️ A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km hr^-1 in 10 minutes . Find its acceleration in the units of ms^-2 .

    A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km hr

    Question βˆ’1

    in 10 minutes. Find its acceleration in the units of ms

    βˆ’2 .

    A

    4

    B

    0.067

    C

    0.0185

    D

    1.11

    Medium Open in App

    Updated on : 2022-09-05

    Solution Verified by Toppr

    Correct option is C)

    Given: t=10 min=10Γ—60=600s

    Initial speed of train is given as: u=0 ms

    βˆ’1

    Final speed of train is given as: v=40 km h

    βˆ’1 = 3600 40Γ—1000 ​ =11.1 ms βˆ’1

    Now acceleration is given by the relation:

    a= t vβˆ’u ​ = 600 11.1βˆ’0 ​ =0.0185 ms βˆ’2

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    A train starting from a railway station and moving with uniform acceleration attains a speed of 40 kmh 1 in 10 minutes. Find its acceleration in the units of ms 2.

    A train starting from a railway station and moving with uniform acceleration attains a speed of 40 kmh 1 in 10 minutes. Find its acceleration in the units of ms 2.

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    A train starting from a railway station and moving with uniform acceleration attains a speed of 40 kmh 1 in 10 minutes. Find its acceleration in the units of ms 2.

    Question

    A train starting from a railway station and moving with uniform acceleration attains a speed of

    40 kmh-1 in 10 minutes

    . Find its acceleration in the units of

    ms-2 . Open in App Solution

    Step 1: Given

    Initial speed, u = 0 ms-2 Final velocity,

    v = 40 kmh-1 = 40Γ—518=1009ms-1

    Time,

    t = 10minutes = 600 seconds

    Step 2: Formula used and calculation

    Using the first equation of motion

    v = u +at

    β‡’1009ms-1 = 0 + 600 sΓ—a

    β‡’a = 1009Γ—600 ms-2 β‡’a = 0.0185 ms-2

    Hence, the acceleration of the train is

    0.0185 ms-2 . Suggest Corrections 150

    SIMILAR QUESTIONS

    Q.

    A train starting from a railway station and moving with uniform acceleration attains a speed 40 km hβˆ’1 in 10 minutes. Find its acceleration.

    Q. A train starting from a railway station and moving with uniform acceleration attains a speed of

    40 k m h r βˆ’ 1 in 10 m i n u t e s

    . Find its acceleration in the units of

    m s βˆ’ 2 .

    Q. Question 3

    A train starting from a railway station and moving with uniform acceleration attains a speed 40 kmph in 10 minutes. Find its acceleration.

    Q.

    a train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes . find its accerelation

    Q.

    A train starting from a railway station and moving with uniform acceleration, attains a speed of 40 km/h in 10 minutes. Its acceleration is :

    ΰ€Έΰ₯ΰ€°ΰ₯‹ΰ€€ : byjus.com

    A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km ^(

    Here, initial speed (u) = 0 ms ^(-1), t = 10 minute =600s Final velocity (v) = 40 km h ^(-1) =(40 xx 1000)/(3600) = (100)/(9) ms ^(-1) Acceleration = (v0u)/(t)= (( 100)/(9) - 0)/( 600) = (100)/( 9 xx 600) = (1)/( 54) m s^(-1) The acceleration is (1)/(54) ms ^(-2).

    Home > English > Class 9 > Physics > Chapter > Motion >

    A train starting from a railwa...

    A train starting from a railway station and moving with uniform acceleration attains a speed of

    40k m βˆ’1 40km-1

    in 10 minute. Find its acceleration.

    Updated On: 27-06-2022

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    Text Solution Open Answer in App Solution Here, initial speed (u)=0m s βˆ’1 , (u)=0ms-1, t=10 t=10 minute =600s =600s Final velocity (v)=40km h βˆ’1 = 40Γ—1000 3600 = 100 9 m s βˆ’1

    (v)=40kmh-1=40Γ—10003600=1009ms-1

    Acceleration = v0u t = 100 9 βˆ’0 600 = 100 9Γ—600 = 1 54 m s βˆ’1

    =v0ut=1009-0600=1009Γ—600=154ms-1

    The acceleration is 1 54 m s βˆ’2 . 154ms-2. Answer

    Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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    Mohammed 15 day ago
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