General guidance

Concepts and reason

The concepts required to solve this problem are position (or displacement), velocity, and acceleration.

Acceleration and position can be derived from the knowledge of velocity of the particle. Acceleration is defined as the rate of change in velocity (that is, derivative of velocity gives acceleration), and reversibly integration of acceleration would give velocity.

Velocity is defined as the rate of change in displacement (that is, derivative of displacement gives velocity), thus integration of velocity would give displacement. Thus, acceleration and position of the particle is calculated from integration of acceleration and velocity respectively.

Fundamentals

Acceleration of the particle is expressed as,

$a = \frac{{dv}}{{dt}}$

Here $a$ is acceleration, $\frac{{dv}}{{dt}}$ is time $\left( t \right)$ derivative of velocity $\left( v \right)$ .

Velocity of the particle is expressed as,

$v = \frac{{ds}}{{dt}}$

Here $v$ is the velocity, $\frac{{ds}}{{dt}}$ is time $\left( t \right)$ derivative of displacement $\left( s \right)$ .

To solve for displacement in the above equation, integrate on both sides as,

$\begin{array}{c}\\ds = v dt\\\\\int {ds} = \int {v dt} \\\\s = \int {v dt} \\\end{array}$

A vector (say $\vec A$ ) in three-dimensional space is expressed as,

$\vec A = x\hat i + y\hat j + z\hat k$

Here $x, y {\rm{and}} z$ denote the magnitude of vector $\vec A$ in $x, y {\rm{and}} z$ direction, respectively and $\hat i,{\rm{ }}\hat j{\rm{ and }}\hat k$ are unit vectors in the direction of $x, y {\rm{and}} z$ , respectively.

2-norm (or) magnitude of $\vec A$ (given above) is computed as,

$\left\| {\vec A} \right\| = \sqrt {{x^2} + {y^2} + {z^2}}$

Magnitude of a vector (say $\vec A$ ) is sometimes also represented as $\left| {\vec A} \right|$ .

General sign convection for axis: Distance along the axis is taken as positive and that opposite to the axis is taken as negative.

First Step | All Steps | Answer Only

Step-by-step

Step 1 of 2

Write the expression of acceleration as,

$a = \frac{{dv}}{{dt}}$

Substitute $\left[ {\left( {16{t^2}} \right)\hat i + \left( {4{t^3}} \right)\hat j + \left( {5t + 2} \right)\hat k} \right]\frac{{\rm{m}}}{{\rm{s}}}$ for $v$ , and solve for acceleration as,

$\begin{array}{c}\\\vec a = \frac{d}{{dt}}\left[ {\left( {16{t^2}} \right)\hat i + \left( {4{t^3}} \right)\hat j + \left( {5t + 2} \right)\hat k} \right]\frac{{\rm{m}}}{{\rm{s}}}\\\\ = \left[ {\left( {32t} \right)\hat i + \left( {12{t^2}} \right)\hat j + \left( 5 \right)\hat k} \right]\frac{{\rm{m}}}{{{{\rm{s}}^2}}}\\\end{array}$

Now to determine the acceleration at a given time, substitute the given value of time in the equation of acceleration obtained above and solve,

Thus, at $t = 2 {\rm{s}}$ ,

$\begin{array}{c}\\\vec a = \left[ {\left( {32\left( {2 {\rm{s}}} \right)} \right)\hat i + \left( {12{{\left( {2 {\rm{s}}} \right)}^2}} \right)\hat j + \left( 5 \right)\hat k} \right]\frac{{\rm{m}}}{{{{\rm{s}}^2}}}\\\\ = \left[ {64 \hat i + 48\hat j + 5 \hat k} \right]\frac{{\rm{m}}}{{{{\rm{s}}^2}}}\\\end{array}$

Now calculate the magnitude of acceleration vector, $\vec a$ .

$\left\| {\vec a} \right\| = \sqrt {{x^2} + {y^2} + {z^2}}$

Substitute $64$

The velocity of a particle is given by v={16t^2i + 4t^3j + (5t +2)k} m/s, where t is in seconds. If the particle is at the origin when t=0, determine the magnitude of the particle's acceleration when t=2s. Also, what is the x,y,z coordinate position of the particle at this instant

Public Answer