The position of a particle moving along an x axis is given by x = 17 t 2 - 2.0 t

Question

The position of a particle moving along an x axis is given by x = 17t2 - 2.0t3, where x is in meters and t is in seconds.

99

48

-2

182

5.67

48.2

2.83

-34

33

The position of a particle moving along an x axis is given by x = 17t2 - 2.0t3, where x is in meters and t is in seconds. Determine the position, velocity, and acceleration of the particle at t = 3.0 s. What is the maximum positive coordinate reached by the particle? What is the maximum positive velocity reached by the particle? What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? Determine the average velocity of the particle between t = 0 and t = 3 s.

Explanation / Answer

x=17t^2-2t^3

v=dx/dt=34t-6t^2

a=dv/dt=34-12t.

a)so at t=3s,

x=99m,v=48m/s,-2m/s^2.

b)at max. positive coorinate dx/dt=0

So,t=34/6s.

So x(max)=181.96m.

c)for max positive vel ,dv/dt=0

So.t=34/12s

So V(max)=48.17m/s.

d)particle is not movig at t=34/6 s

a=-34m/s^2

e)at t=0 x=0; at t=3 x=99m

SO av. vel=[x(3)-x(0)]/3=99/3=33m/s

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