Question

A particle travels along a straight line with a velocity v=(12-3t²) m/s, where t is in seconds. When t=1s, the particle is located 10m to the left of the origin. Determine the acceleration when t=4s, the displacement from t=0 to t=10s, the distance the particle travels during this time period.

Answer 1

Concepts and reason

Distance is defined the total path covered by a body when it is under motion.

Displacement is defined as the minimum distance between the initial and final position of the object travelled by it.

Velocity is defined as the rate of change of displacement of the body with time. It defines how faster a body is moving.

Acceleration is defined as the rate of change of velocity with time. It defines whether the body is speeding up or down.

Fundamentals

‎Suppose the equation for displacement of the body is given as a function of time, that is $s(t)$ .

The velocity is given as:

$v(t) = \frac{{ds(t)}}{{dt}}$

And the acceleration is given as:

$\begin{array}{c}\\a(t) = \frac{{{d^2}s(t)}}{{d{t^2}}}\\\\ = \frac{{dv(t)}}{{dt}}\\\end{array}$

Also, if the velocity of the body is given in terms of time, the equation for displacement is given as:

$\begin{array}{l}\\\frac{{ds(t)}}{{dt}} = v(t)\\\\s(t) = \int {v(t)} dt + C\\\end{array}$

Here, C is the constant of integration and calculated by boundary equations.

The velocity of the particle traveling along the straight line is:

Express the acceleration of the particle in terms of t.

Differentiate the equation for velocity with respect to time t.

Substitute $\left( {12 - 3\,{t^{\,2}}} \right){\rm{m/s}}$ for v.

Calculate the acceleration of the particle at $t=4s $.

Substitute $4\,{\rm{s}}$ for t.

Negative sign indicates that the particle is deaccelerating.

The velocity of the particle traveling along the straight line is:

Express the position of the particle in terms of time t.

$\frac{{ds(t)}}{{dt}} = v(t)$

Integrate the above equation.

$s(t) = \int {v(t)} dt + C$

Here, C is the constant of integration and calculated by boundary equations.

Substitute $\left( {12 - 3\,{t^{\,2}}} \right){\rm{m/s}}$ for v.

$\begin{array}{l}\\s(t) = \int {\left( {12 - 3\,{t^{\,2}}} \right)} dt + C\\\\s = 12t - {t^3} + C\\\end{array}$ …… (1)

Use boundary conditions:

At time $t = 1{\rm{ s}}$ , the particle is at 10 m to the left.

Substitute $t = 1$ and $s = - 10{\rm{ m}}$ .

$\begin{array}{l}\\ - 10 = 12 \times 1 - {1^3} + C\\\\C = - 21\\\end{array}$

Substitute -21 for C in equation (1).

$s = 12t - {t^3} - 21$ …… (2)

Calculate the position of the particle at $t = 1{\rm{ s}}$ .

$\begin{array}{c}\\s\left( 0 \right) = 12 \times 0 - {0^3} - 21\\\\ = - 21{\rm{ m}}\\\end{array}$

Calculate the position of the particle $t = 10{\rm{ s}}$ .

$\begin{array}{c}\\s\left( {10} \right) = 12 \times 10 - {10^3} - 21\\\\ = - 901{\rm{ m}}\\\end{array}$

Calculate the displacement from $t = 1{\rm{ s}}$ to $t = 10{\rm{ s}}$ .

$\Delta s = s(10) - s(0)$

Substitute $- 901{\rm{ m}}$ for $s\left( {10} \right)$ and $- 21{\rm{ m}}$ for $s\left( 0 \right)$ .

$\begin{array}{c}\\\Delta s = - 901 - \left( { - 21} \right)\\\\ = - 880{\rm{ m}}\\\end{array}$

Negative sign indicates that the particle is displaced to the left side of the origin.

To calculate the distance travelled by the particle, calculate the point of zero velocity.

The velocity of the particle traveling along the straight line is:

Put the velocity equal to zero.

$\begin{array}{l}\\0 = \left( {12 - 3\,{t^{\,2}}} \right)\\\\3{t^2} = 12\\\\{t^2} = 4\\\\t = 2{\rm{ s}}\\\end{array}$

Therefore, the velocity at time equal 2 s is zero.

Calculate the position of the particle at $t = 2{\rm{ s}}$ .

Substitute $2{\rm{ s}}$ for t in equation (2).

$\begin{array}{c}\\s(2) = 12 \times 2 - {2^3} - 21\\\\ = - 5{\rm{ m}}\\\end{array}$

Calculate the distance travelled by particle from $t = 1{\rm{ s}}$ to $t = 10{\rm{ s}}$ .

${s_d} = \left[ {s\left( 2 \right) - s\left( 0 \right)} \right] + \left[ {s\left( 2 \right) - s\left( {10} \right)} \right]$

Substitute $- 901{\rm{ m}}$ for $s\left( {10} \right)$ , $- 5{\rm{ m}}$ for $s(2)$ and $- 21{\rm{ m}}$ for $s\left( 0 \right)$ .

$\begin{array}{c}\\{s_d} = \left[ { - 5 - \left( { - 21} \right)} \right] + \left[ {\left( { - 5} \right) - \left( { - 901} \right)} \right]\\\\ = 912{\rm{ m}}\\\end{array}$

Ans:

The acceleration of the particle at $t=4s $ is $- \,24\,{\rm{m/}}{{\rm{s}}^{\,{\rm{2}}}}$ .

The displacement from $t = 1{\rm{ s}}$ to $t = 10{\rm{ s}}$ is $- 880{\rm{ m}}$ .

The distance the particle from period $t = 1{\rm{ s}}$ to $t = 10{\rm{ s}}$ is $912{\rm{ m}}$ .