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constant velocityconstant accelerationvariable velocity variable acceleration

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BTranscript

Time | Transcript |
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00:00 - 00:59 | question says that A particle moves along the parabola y square is equal to 2 a x in such a way that its projection on why has a constant velocity equation of y axis has a constant velocity given to us we have to show that its projection on x-axis moves with constant acceleration so that accesses acceleration will be constantly constantly have to prove this should parabola is y square is equal to 2 X Suresh first differentiate with respect to X respectively so as to why Divya Pondatti is equal to 2 a s e t f deficit with respect to x now from here Divya Pondatti is equal to |

01:00 - 01:59 | a y y x dx upon DT no again differentiate with respect to X with respect to teach sorry it is due to buy a phone square is equal to is constant taken outside 1 taking on my way constantly first and definition of this is due to X upon X square + now taking DX upon duty constant definition of 1 by x minus 1 upon y square now from here it is given that the creation on y axis has a constant velocity is constant velocity this will be equal to zero because if it is equal to zero is equal equal to zero so from here we get as one upon Y to X upon DT |

02:00 - 02:59 | is equal to minus one upon y square upon DT this private cancel this shirt is equal to 2 X upon x square is equal to minus one upon by DX upon DT Nokia rating value of the second ODI from this equation that is the 2 X upon x square is equal to minus 1 by why density value is why Upon A upon DT I will cancel by this so what did from hair d2x upon DT square is equal to minus one upon a into a pond ET now this is given constant in the question of y axis and age |

03:00 - 03:59 | Austin town is already a constant term constant and this is our acceleration at x-axis so that access is the acceleration is completely constant we have proved it question |