# 1.15 Rectilinear motion  (Page 2/6)

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This attribute of rectilinear motion allows us to do away with the need to use vector notation and vector algebra for quantities with directional attributes like position vector, displacement and velocity. Instead, the vectors are treated simply as scalars with one qualification that vectors in the direction of chosen reference is considered positive and vectors in the opposite direction to the chosen reference is considered negative.

The most important aspect of the sign convention is that a vector like velocity can be expressed by a scalar value say, 5 m/s. Though stated without any aid for specifying direction like using unit vector, the direction of the velocity is indicated, which is in the positive x-direction. If the velocity of motion is -5 m/s, then the velocity is in the direction opposite to the direction of reference.

To illustrate the construct, let us consider a motion of a ball which transverses from O to A to B to C to O along x-axis as shown in the figure.

The velocities at various points of motion in m/s (vector form) are :

$\begin{array}{l}{\mathbf{v}}_{\mathbf{O}}=2\mathbf{i};\phantom{\rule{4pt}{0ex}}{\mathbf{v}}_{\mathbf{A}}=3\mathbf{i};\phantom{\rule{4pt}{0ex}}{\mathbf{v}}_{\mathbf{B}}=-4\mathbf{i};\phantom{\rule{4pt}{0ex}}{\mathbf{v}}_{\mathbf{C}}=3\mathbf{i}\end{array}$

Going by the scalar construct, we can altogether drop use of unit vector like " i " in describing all vector quantities used to describe motion in one dimension. The velocities at various points of motion in m/s (in equivalent scalar form) can be simply stated in scalar values for rectilinear motion as :

$\begin{array}{l}{v}_{O}=2;\phantom{\rule{4pt}{0ex}}{v}_{A}=3;\phantom{\rule{4pt}{0ex}}{v}_{B}=-4;\phantom{\rule{4pt}{0ex}}{v}_{C}=3\end{array}$

Similarly, we can represent position vector simply by the component in one direction, say x, in meters, as :

$\begin{array}{l}{x}_{O}=0;\phantom{\rule{4pt}{0ex}}{x}_{A}=5;\phantom{\rule{4pt}{0ex}}{x}_{B}=10;\phantom{\rule{4pt}{0ex}}{x}_{C}=-5\end{array}$

Also, the displacement vector (in meters) is represented with scalar symbol and value as :

$\begin{array}{l}\mathrm{OA}=5;\phantom{\rule{4pt}{0ex}}\mathrm{OB}=10;\phantom{\rule{4pt}{0ex}}\mathrm{OC}=5\end{array}$

Following the same convention, we can proceed to write defining equations of speed and velocity in rectilinear motion as :

$\begin{array}{l}|v|=|\frac{đx}{đt}|\end{array}$

and

$\begin{array}{l}v=\frac{đx}{đt}\end{array}$

## Rectilinear motion

Problem : If the position of a particle along x – axis varies in time as :

$\begin{array}{l}x=2{t}^{2}-3t+1\end{array}$

Then :

• What is the velocity at t = 0 ?
• When does velocity become zero?
• What is the velocity at the origin ?
• Plot position – time plot. Discuss the plot to support the results obtained for the questions above.

Solution : We first need to find out an expression for velocity by differentiating the given function of position with respect to time as :

$\begin{array}{l}v=\frac{đ}{đt}\left(2{t}^{2}-3t+1\right)=4t-3\end{array}$

(i) The velocity at t = 0,

$\begin{array}{l}v=4x0-3=-3\mathrm{m/s}\end{array}$

(ii) When velocity becomes zero :

For v = 0,

$\begin{array}{l}4t-3=0\\ ⇒t=\frac{3}{4}=\mathrm{0.75 s.}\end{array}$

(iii) The velocity at the origin :

At origin, x = 0,

$\begin{array}{l}x=2{\mathbf{t}}^{2}-3t+1=0\\ ⇒2{\mathbf{t}}^{2}-2t-t+1=0\\ ⇒2t\left(t-1\right)-\left(t-1\right)=0\\ ⇒t=\mathrm{0.5 s, 1 s.}\end{array}$

This means that particle is twice at the origin at t = 0.5 s and t = 1 s. Now,

$\begin{array}{l}v\left(t=\mathrm{0.5 s}\right)=4t-3=4x0.5-3=\mathrm{-1 m/s.}\end{array}$

Negative sign indicates that velocity is directed in the negative x – direction.

$\begin{array}{l}v\left(t=\mathrm{1 s}\right)=4t-3=4x1-3=\mathrm{1 m/s.}\end{array}$

We observe that slope of the curve from t = 0 s to t<0.75 s is negative, zero for t = 0.75 and positive for t>0.75 s. The velocity at t = 0, thus, is negative. We can realize here that the slope of the tangent to the curve at t = 0.75 is zero. Hence, velocity is zero at t = 0.75 s.

The particle arrives at x = 0 for t = 0.5 s and t = 1 s. The velocity at first arrival is negative as the position falls on the part of the curve having negative slope, whereas the velocity at second arrival is positive as the position falls on the part of the curve having positive slope.

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