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Question: The position of a particle is given by \(r = 3.0t\widehat i + 2.0{t^2}\widehat j + 5.0\widehat k\) ,...

The position of a particle is given by r=3.0ti^+2.0t2j^+5.0k^r = 3.0t\widehat i + 2.0{t^2}\widehat j + 5.0\widehat k , where ‘tt’ is in second and the coefficients have the proper units for ‘rr’to be in metre. Find u(t)u(t) and a(t)a(t) of the particle.

Explanation

Solution

Hint We know that velocity is equal to rate of change of position with respect to time.
i.e., v=drdt\overrightarrow v = \dfrac{{d\overrightarrow r }}{{dt}}
where, r\overrightarrow r is position vector.
Now, we know acceleration is the rate of change of velocity with respect to time.
i.e., a=dvdt\overrightarrow a = \dfrac{{d\overrightarrow v }}{{dt}}
where, v\overrightarrow v is velocity vector.

Complete Step by step solution
Given: the position vector = r=3.0ti^+2.0t2j^+5.0k^r = 3.0t\widehat i + 2.0{t^2}\widehat j + 5.0\widehat k
We know velocity is the rate of change of position with respect to time.
Therefore, we have
u(t)=drdt u(t)=d(3.0ti^+2.0t2j^+5.0k^)dt u(t)=3.0i^+4tj^+0 u(t)=3.0i^+4tj^  u(t) = \dfrac{{dr}}{{dt}} \\\ u(t) = \dfrac{{d\left( {3.0t\widehat i + 2.0{t^2}\widehat j + 5.0\widehat k} \right)}}{{dt}} \\\ u(t) = 3.0\widehat i + 4t\widehat j + 0 \\\ u(t) = 3.0\widehat i + 4t\widehat j \\\
Hence, the required velocity of the particle is u(t)=3.0i^+4tj^u(t) = 3.0\widehat i + 4t\widehat j
Now we know that the acceleration is rate of change of velocity.
Therefore, we have
a(t)=du(t)dta(t) = \dfrac{{du(t)}}{{dt}}
Using the above value of u(t)u(t) we get
a(t)=d(3.0i^+4tj^)dt a(t)=0+4j^ a(t)=4j^  a(t) = \dfrac{{d(3.0\widehat i + 4t\widehat j)}}{{dt}} \\\ a(t) = 0 + 4\widehat j \\\ a(t) = 4\widehat j \\\

Hence, the required value of acceleration is a(t)=4j^a(t) = 4\widehat j

Note The case of constant acceleration and the motion in a straight line yields some simple equation that permits the evaluation of the velocity and the position of the vehicle if the initial conditions are known. From the definition, we know a=dvdta = \dfrac{{dv}}{{dt}} , the velocity at later time tt can be determined from the initial velocity, v(0)v(0) , and the constant acceleration aa , by integration. This gives
v(t)=v(0)+atv(t) = v(0) + at
Similarly, on implementing different conditions of position and velocity we get
s(t)=v(0)+12at2 v2(t)=v2(0)+2as  s(t) = v(0) + \dfrac{1}{2}a{t^2} \\\ {v^2}(t) = {v^2}(0) + 2as \\\