### Discussion :: Applied Mechanics - Section 1 (Q.No.4)

Saurabh said: (Jul 22, 2013) | |

x = t^{3} - 3t^{2} + 5.dx/dt = t ^{2}/2 - 3t.= 12.5 - 15. = -2.5. 4.5 - 9. = -4.5. 4.5/2.5 = 1.8 = 2. |

Jadecliff said: (Jul 23, 2013) | |

x = Distance. v = Velocity = dx/dt. a = Acceleration = dv/dt. From x = t ^{3} - 3t^{2} + 5.v = 3t ^{2} - 6t.a = 6t - 6. By substituting 5 sec and 3 sec in a. a1 = 24. a2 = 12. a1 : a2 = 2 : 1. |

Dawit Tamiru said: (Nov 20, 2014) | |

Another method d"x/dt = 6t-6. 6(5)-6 = 24 and 6(3)-6 = 12. we got the ratio 2. |

Priyanka Marmat said: (Feb 16, 2015) | |

d"x/dt = 6t-6. 6(5)-6 = 24. 6(4)-6 = 12. 24/12 = 2. The ratio is 2. |

Abhay said: (Jan 25, 2016) | |

Equation of motion x = t^3-3t^2+5. Equation to calculate distance (in question it is given that the particle is moving along a straight line). Differentiate with time to get velocity: v = dx/dt = 3t^2-6t.....Velocity. a = d^2x/dt = 6t-6.....Acceleration. Hence, when t = 5 sec. a = 6*5 - 6 = 24. When t = 3sec. a = 6*3 - 6 = 12. Ratio = 24/2 = 2. |

Tariku Teshome said: (Jul 18, 2016) | |

Answer for; d = x = distance. v = velocity. a = acceleration. 11, 5 & 2 respectively. |

Laxmipriyadhal12345@Gmail said: (Nov 11, 2016) | |

Thank you @Abhay. |

Abhinandan said: (Apr 23, 2017) | |

@Jadecliff and @Abhay. Thanks for giving the step by step solution. |

Engr.Mohbat Baloch said: (Sep 10, 2017) | |

x = t3 - 3t2 + 5. dx/dt = t2/2 - 3t. putting t=5 = 12.5 - 15. = -2.5. putting t=3 4.5 - 9. = -4.5. 4.5/2.5 = 1.8 ~2. |

Eric said: (Oct 16, 2018) | |

Thanks @Abhay. It is easy to understand it. |

Sekhar Alvakonda said: (Dec 14, 2018) | |

Thank you all. |

Sheraz said: (Aug 11, 2021) | |

Thanks for explaining. |

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