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Question: The position of a particle is given by \(r = 3t\hat i - 2{t^2}\hat j + 4\hat k\,m\) where \(t\) is i...

The position of a particle is given by r=3ti^−2t2j^+4k^ mr = 3t\hat i - 2{t^2}\hat j + 4\hat k\,m where tt is in seconds and the coefficients have the proper units for rr to be in meters. (a) Find the velocity and acceleration of the particle. (b) What is the magnitude and direction of the velocity of the particle at t=2 st = 2\,s?

Explanation

Solution

Differentiate the position vector with respect to time once to find the velocity vector and twice to find the acceleration vector. To find the velocity vector at any given time, substitute the value of time in it. The direction of the velocity will be the angle formed by the velocity vector with the x-axis.
Formula used:
⇒v⃗=dr⃗dt\Rightarrow \vec v = \dfrac{{d\vec r}}{{dt}} where v⃗\vec v is the velocity vector
⇒a⃗=dv⃗dt\Rightarrow \vec a = \dfrac{{d\vec v}}{{dt}} where a⃗\vec a is the acceleration vector

Complete step by step solution:
The velocity of any particle is the rate of change of its position with respect to time and the acceleration is in turn equal to the rate of change of velocity with respect to time.
(a) So, we can calculate the velocity of the particle as
⇒v⃗=dr⃗dt\Rightarrow \vec v = \dfrac{{d\vec r}}{{dt}}
⇒v⃗=ddt(3ti^−2t2j^+4k^ )\Rightarrow \vec v = \dfrac{d}{{dt}}(3t\hat i - 2{t^2}\hat j + 4\hat k\,)
On differentiating each component individually, we get
⇒v⃗=3i^−4tj^\Rightarrow \vec v = 3\hat i - 4t\hat j
Now that we know the velocity vector, we can calculate the acceleration vector using
⇒a⃗=dv⃗dt\Rightarrow \vec a = \dfrac{{d\vec v}}{{dt}}
⇒a⃗=ddt(3i^−4tj^)\Rightarrow \vec a = \dfrac{d}{{dt}}(3\hat i - 4t\hat j)
Again, differentiating each component separately, we can obtain the acceleration vector as
⇒a⃗=−4j^\Rightarrow \vec a = - 4\hat j
(b) To find the magnitude and direction of the velocity of the particle at t=2 s, we place the value of t in
⇒v⃗=3i^−4tj^\Rightarrow \vec v = 3\hat i - 4t\hat j
Which gives us
⇒v⃗=3i^−8j^\Rightarrow \vec v = 3\hat i - 8\hat j
The magnitude of the velocity is calculated as:
⇒∣v∣ = 32+82\Rightarrow |v|\, = \,\sqrt {{3^2} + {8^2}}
⇒∣v∣ = 73=8.54 m/s\Rightarrow |v|\, = \,\sqrt {73} = 8.54\,m/s
Since the velocity component has only x and y components, the direction of the velocity vector with the x-axis can be calculated from the ratio of velocity in the y-direction to the velocity in the x-direction as:
⇒θ=tan⁡−1(−83)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{ - 8}}{3}} \right)
⇒θ=−69.44∘\Rightarrow \theta = - 69.44^\circ with the x-axis. Hence the velocity vector has magnitude ∣v∣ =8.54 m/s|v|\, = 8.54\,m/s and lies at an angle of θ=−69.44∘\theta = - 69.44^\circ from the x-axis.

Note:
While dealing with derivatives of vectors, we must be careful in differentiating the individual components of the vectors while keeping their directions as it is. But since the vectors themselves change as a function of time, their magnitude and directions will also change with time.