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Find the time t0 when x-coordinate of th...

Find the time `t_0` when x-coordinate of the particle is zero.

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The correct Answer is:
B, C, D
`8.25 s`
(b) Total displacement =`4+4-16-32-16=-56 m`
This is also equal to `x_f-x_i=-46-10=-56 m`
Total distance=`4+4+16+32+16=72 m`
Total time =`16s`
Now, average velocity=total displacement/total time
` = -56/16=-3.5 m//s`
average speed =`("total distance")/("total time")`
`=72/16=4.5 m//s`
(c) Average acceleration=`(Delta v)/(Delta t)`
`=(v_f-v_i)/(Delta t)=(v_(8 sec)-v_(2 sec))/(8-2)`
`=(-8-4)/6=-2 m//s^2`
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Knowledge Check

  • Velocity of particle moving along positive x-direction is v = (40-10t)m//s . Here,t is in seconds. At time t=0, tha x coordinate of particle is zero. Find the time when the particle is at a distance of 60 m from origin.

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    D
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  • If v = x^2 - 5x + 4 , find the acceleration of the particle when velocity of the particle is zero.

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    D
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  • Find the time when the displacement of the particle becomes zero. Assume velocity and position at t = 0 to be and - 10 m respectively.

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