General
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Physics-

A train moves from one station to another 2 hours time. Its speed-time graph during this motion is shown in the figure. The maximum acceleration during the journey is

Physics-General

  1. 140 blank k m blank h to the power of negative 2 end exponent    
  2. 160 blank k m blank h to the power of negative 2 end exponent    
  3. 100 blank k m blank h to the power of negative 2 end exponent    
  4. 120 blank k m blank h to the power of negative 2 end exponent    

    Answer:The correct answer is: 160 blank k m blank h to the power of negative 2 end exponentMaximum acceleration will be represented by C D part of the graph
    Acceleration equals fraction numerator d v over denominator d t end fraction equals fraction numerator left parenthesis 60 minus 20 right parenthesis over denominator 0.25 end fraction equals 160 blank k m divided by h to the power of 2 end exponent

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    General
    physics-

    In the given v minus t graph, the distance travelled by the body in 5 will be

    Area between v minus t graph and time-axis
    equals fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 20 plus 3 cross times 20 plus fraction numerator 1 over denominator 2 end fraction cross times 1 cross times 20 plus fraction numerator 1 over denominator 2 end fraction cross times 1 cross times 20
    equals 100 m.

    In the given v minus t graph, the distance travelled by the body in 5 will be

    physics-General
    Area between v minus t graph and time-axis
    equals fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 20 plus 3 cross times 20 plus fraction numerator 1 over denominator 2 end fraction cross times 1 cross times 20 plus fraction numerator 1 over denominator 2 end fraction cross times 1 cross times 20
    equals 100 m.
    General
    physics-

    A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t subscript 1 end subscript and t subscript 2 end subscript seconds respectively, then

    If a stone is dropped from height h then
    h equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent(i)
    if a stone is thrown upward with velocity u then
    h equals negative u blank t subscript 1 end subscript plus fraction numerator 1 over denominator 2 end fraction g t subscript 1 end subscript superscript 2 end superscript(ii)
    If a stone is thrown downward with velocity u then
    h equals u t subscript 2 end subscript plus fraction numerator 1 over denominator 2 end fraction g t subscript 2 end subscript superscript 2 end superscript(iii)
    From (i) (ii) and (iii) we get
    negative u t subscript 1 end subscript plus fraction numerator 1 over denominator 2 end fraction g t subscript 1 end subscript superscript 2 end superscript equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent(iv)
    u t subscript 2 end subscript plus fraction numerator 1 over denominator 2 end fraction g t subscript 2 end subscript superscript 2 end superscript equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent(v)
    Dividing (iv) and (v) we get
    therefore fraction numerator negative u t subscript 1 end subscript over denominator u t subscript 2 end subscript end fraction equals fraction numerator fraction numerator 1 over denominator 2 end fraction g left parenthesis t to the power of 2 end exponent minus t subscript 1 end subscript superscript 2 end superscript right parenthesis over denominator fraction numerator 1 over denominator 2 end fraction g left parenthesis t to the power of 2 end exponent minus t subscript 2 end subscript superscript 2 end superscript right parenthesis end fraction
    Or negative fraction numerator t subscript 1 end subscript over denominator t subscript 2 end subscript end fraction equals fraction numerator t to the power of 2 end exponent minus t subscript 1 end subscript superscript 2 end superscript over denominator t to the power of 2 end exponent minus t subscript 2 end subscript superscript 2 end superscript end fraction
    By solvingblank t equals square root of t subscript 1 end subscript t subscript 2 end subscript end root

    A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t subscript 1 end subscript and t subscript 2 end subscript seconds respectively, then

    physics-General
    If a stone is dropped from height h then
    h equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent(i)
    if a stone is thrown upward with velocity u then
    h equals negative u blank t subscript 1 end subscript plus fraction numerator 1 over denominator 2 end fraction g t subscript 1 end subscript superscript 2 end superscript(ii)
    If a stone is thrown downward with velocity u then
    h equals u t subscript 2 end subscript plus fraction numerator 1 over denominator 2 end fraction g t subscript 2 end subscript superscript 2 end superscript(iii)
    From (i) (ii) and (iii) we get
    negative u t subscript 1 end subscript plus fraction numerator 1 over denominator 2 end fraction g t subscript 1 end subscript superscript 2 end superscript equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent(iv)
    u t subscript 2 end subscript plus fraction numerator 1 over denominator 2 end fraction g t subscript 2 end subscript superscript 2 end superscript equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent(v)
    Dividing (iv) and (v) we get
    therefore fraction numerator negative u t subscript 1 end subscript over denominator u t subscript 2 end subscript end fraction equals fraction numerator fraction numerator 1 over denominator 2 end fraction g left parenthesis t to the power of 2 end exponent minus t subscript 1 end subscript superscript 2 end superscript right parenthesis over denominator fraction numerator 1 over denominator 2 end fraction g left parenthesis t to the power of 2 end exponent minus t subscript 2 end subscript superscript 2 end superscript right parenthesis end fraction
    Or negative fraction numerator t subscript 1 end subscript over denominator t subscript 2 end subscript end fraction equals fraction numerator t to the power of 2 end exponent minus t subscript 1 end subscript superscript 2 end superscript over denominator t to the power of 2 end exponent minus t subscript 2 end subscript superscript 2 end superscript end fraction
    By solvingblank t equals square root of t subscript 1 end subscript t subscript 2 end subscript end root
    General
    maths-

    ‘0’ is the centre of the circle and B A C plus B O C equals 120 to the power of ring operator end exponent text  then  end text angle C O B equals

    ‘0’ is the centre of the circle and B A C plus B O C equals 120 to the power of ring operator end exponent text  then  end text angle C O B equals

    maths-General
    General
    maths-

    Find the value of ‘x’ in the given figure

    Find the value of ‘x’ in the given figure

    maths-General
    General
    maths-

    In the figure shown, PT and PAB are the tangent and the secant drawn to a circle. If PT = 12 cm and PB = 8 cm then AB is

    In the figure shown, PT and PAB are the tangent and the secant drawn to a circle. If PT = 12 cm and PB = 8 cm then AB is

    maths-General
    General
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    The length of AC in the figure

    The length of AC in the figure

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    General
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    In the following circle, ‘O’ is the centre .If C A B equals 35 to the power of ring operator end exponent text  then  end text left floor A B C equals

    In the following circle, ‘O’ is the centre .If C A B equals 35 to the power of ring operator end exponent text  then  end text left floor A B C equals

    maths-General
    General
    maths-

    In the given diagram fraction numerator A O over denominator O C end fraction equals fraction numerator B O over denominator O D end fraction equals fraction numerator 1 over denominator 2 end fraction and A B equals 5 c m. Find the value of DC.

    In the given diagram fraction numerator A O over denominator O C end fraction equals fraction numerator B O over denominator O D end fraction equals fraction numerator 1 over denominator 2 end fraction and A B equals 5 c m. Find the value of DC.

    maths-General
    General
    maths-

    In the given figure, find the values of x y, and z.

    In the given figure, find the values of x y, and z.

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    General
    maths-

    ABCD is a quadrilateral. AD and BD are the angle bisectors of angle A and B which meet at ‘O’. If C equals 70 to the power of ring operator end exponent comma L equals 50 to the power of ring operator end exponent. text  Find  end text left floor A O B

    ABCD is a quadrilateral. AD and BD are the angle bisectors of angle A and B which meet at ‘O’. If C equals 70 to the power of ring operator end exponent comma L equals 50 to the power of ring operator end exponent. text  Find  end text left floor A O B

    maths-General
    General
    maths-

    In the unit Square, find the distance from E to stack A D with bar on top in terms of a and b, the lengths of DF and AG , respectively.

    In the unit Square, find the distance from E to stack A D with bar on top in terms of a and b, the lengths of DF and AG , respectively.

    maths-General
    General
    maths-

    ABCD is a Parallelogram, A E perpendicular D C and C F perpendicular A D If DC=16cm, AE = 8 cm and CF = 10 cm, Find AD

    ABCD is a Parallelogram, A E perpendicular D C and C F perpendicular A D If DC=16cm, AE = 8 cm and CF = 10 cm, Find AD

    maths-General
    General
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    Two Rectangles ABCD and DBEF are as shown in the figure. The area of Rectangle DBEF is

    Two Rectangles ABCD and DBEF are as shown in the figure. The area of Rectangle DBEF is

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    General
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    In the given figure, ABCD is a Cyclic stack D E with bar on top perpendicular stack A B with bar on top comma B A O equals 40 to the power of ring operator end exponent comma Q A D equals 20 to the power of ring operator end exponent and ∣ O C D equals 50 to the power of ring operator end exponent then A B C equals

    In the given figure, ABCD is a Cyclic stack D E with bar on top perpendicular stack A B with bar on top comma B A O equals 40 to the power of ring operator end exponent comma Q A D equals 20 to the power of ring operator end exponent and ∣ O C D equals 50 to the power of ring operator end exponent then A B C equals

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    General
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    In the figure be low stack A B with bar on top parallel to stack D C with bar on top comma stack A E with rightwards arrow on top parallel to stack B C with bar on top and stack A F with bar on top parallel to stack B D with bar on top If D equals x to the power of ring operator end exponent comma C equals y to the power of 0 end exponent comma open floor E A B equals k to the power of ring operator end exponent close and A B P equals p to the power of ring operator end exponent comma then which of the following is correct ?

    In the figure be low stack A B with bar on top parallel to stack D C with bar on top comma stack A E with rightwards arrow on top parallel to stack B C with bar on top and stack A F with bar on top parallel to stack B D with bar on top If D equals x to the power of ring operator end exponent comma C equals y to the power of 0 end exponent comma open floor E A B equals k to the power of ring operator end exponent close and A B P equals p to the power of ring operator end exponent comma then which of the following is correct ?

    maths-General