Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x"(t) is its acceleration.x(t)=t3−6t2+9t−9,0≤t≤10a) Find the velocity and acceleration of the particle?x′(t)=?x"(t)=?b) Find the open t-intervals on which the particle is moving to the right?c) Find the velocity of the particle when the acceleration is 0?

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Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x&#039;(t) is its velocity, and x&quot;(t) is its acceleration.x(t)=t3−6t2+9t−9,0≤t≤10a) Find the velocity and acceleration of the particle?x′(t)=?x&quot;(t)=?b) Find the open t-intervals on which the particle is moving to the right?c) Find the velocity of the particle when the acceleration is 0?

Clelioo · Accepted Answer

Step 1Position as a function of time is given. We have to find the velocity and acceleration.Recall the famous rule of differentiation:d(xn)/dx=nxn−1Part (a)Velocity =x′(t)=3t2−12t+9Acceleration =x&quot;(t)+6t−12Part (b)When particle is moving to the right, velocity &amp;gt;0Hence, x′(t)&amp;gt;0Hence, 3t2−12t+9&amp;gt;0Or, t2−4t+3&amp;gt;0Or, (t−3)(t−1)&amp;gt;0Hence, t&amp;lt;1 or t&amp;gt;3The bounds for t as stated in question is: 0≤t≤10Hence, the open t-interval will be: [0,1)∪(3,10]Part (c)Acceleration is zero when x&quot;(t)=6t−12=0Hence, t=12/6=2Hence, velocity when acceleration is zero =x′(t=2)=3⋅22−12⋅2+9=−3

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x"(t) is its acceleration

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