### Discussion :: Kinematics of Particle (KOP) - General Questions (Q.No.7)

Rajkiran Hatheshwar said: (Dec 6, 2015) | |

Step 1 - By similar triangle at s=100m, we will get v=(100/3)m/s. Now from s=d/t; find t=3 sec. Then from v=u+at, we will get a=11.11 m/s*s (here u=0). Step 2 - Similarly at s=175m, v=25m/s and t=1sec. So that a=-25m/s*s. Note:- in 2nd case you will not be 0, here you will be 50m/s. THANK YOU. |

Nayan said: (Feb 1, 2017) | |

Please explain how you taking values of t and v? |

Pradeep said: (Feb 27, 2018) | |

@Nayan it is a concept of the similar triangle: which implies 150/50=100/v. Gives v=100/3 and from s=v*t we will get t. |

Raj Shah said: (Nov 25, 2020) | |

Let s0=0m, s1=150m, s2=200m, s3=100m, & s4=175m. a3=[(s3-s0)/(s1-s0)]*v1*[(v1-v0)/(s1-s0)]. a3=11.11m/s*s. a4=[(s4-s1)/(s2-s1)]*v1*[(v2-v1)/(s2-s1)]. a4= -25 m/s*s. |

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