Question

A particle is moving along a straight line and follows given equation y=(t^2-5t+6)m then calculate -
1)Time when particle is at origin
2) Particle's initial position
3) position of particle at time t=4 seconds
4)Particle's initial speed
5) Time when particle is at rest
6) Velocity and speed of particle at time t=2 seconds...​

Answer 1

(i) When particle is at origin, it has no displacement ⇒ y = 0

⇒ y = 0            ⇒ t² - 5t + 6 = 0

⇒ t² - 2t - 3t + 6 = 0

⇒ t(t - 2) - 3(t - 2) = 0

⇒ (t - 2)(t - 3) = 0

 time = 2s       or  t = 3s

Initially, when time = 0,

(ii)  y = t² - 5t + 6

    = 0² - 5(0) + 6

     = 6 m

(iii) Position at t = 4:

y = 4² - 5(4) + 6

   = 16 - 20 + 6

   = 2 m

(iv) speed = |v| = | dy/dt |

         = | d(t² - 5t + 6)/dt |

         = | 2t - 5 |

Initially t = 0,  speed = |2(0) - 5|

                                  = 5 m/s

&  initial velocity = 2(0) - 5 = -5m/s

(v) When it is at rest: v = 0

       2t - 5 = 0

        t = 5/2 = 2.5 s

(vi) At t = 2,

v = 2t - 2  = 2(2) - 2 = 2 m/s

speed = | 2t -2 | = | 2(2) - 2 |

           = 2 m/s

Answer 2

Explanation:

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