2.6 One dimensional motion with constant acceleration  (Page 12/6)

Free falling bodies under gravity represents typical case of motion in one dimension with constant acceleration. A body projected vertically upwards is also a case of constant acceleration in one dimension, but with the difference that body undergoes reversal of direction as well after reaching the maximum height. Yet another set of examples of constant accelerations may include object sliding on an incline plane, motion of an aboject impeded by rough surfaces and many other motions under the influence of gravitational and frictional forces.

The defining differential equations of velocity and acceleration involve only one position variable (say x). In the case of motion under constant acceleration, the differential equation defining acceleration must evaluate to a constant value.

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

and

$$\begin{array}{l}\mathbf{a}=\frac{{đ}^{2}x}{đt^2}\mathbf{i}=k\mathbf{i}\end{array}$$

where k is a positive or negative constant.

The corresponding scalar form of the defining equations of velocity and acceleration for one dimensional motion with constant acceleration are :

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

and

$$\begin{array}{l}a=\frac{{đ}^{2}x}{đt^2}=k\end{array}$$

Constant acceleration

Problem : The position “x” in meter of a particle moving in one dimension is described by the equation :

$$\begin{array}{l}t=\sqrt{x}+1\end{array}$$

where “t” is in second.

• Find the time when velocity is zero.
• Does the velocity changes its direction?
• Locate position of the particle in the successive seconds for first 3 seconds.
• Find the displacement of the particle in first three seconds.
• Find the distance of the particle in first three seconds.
• Find the displacement of the particle when the velocity becomes zero.
• Determine, whether the particle is under constant or variable force.

Solution : Velocity is equal to the first differential of the position with respect to time, while acceleration is equal to the second differential of the position with respect to time. The given equation, however, expresses time, t, in terms of position, x. Hence, we need to obtain expression of position as a function in time.

$$\begin{array}{l}t=\sqrt{x}+1\\ \Rightarrow \sqrt{x}=t-1\end{array}$$

Squaring both sides, we have :

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

This is the desired expression to work upon. Now, taking first differential w.r.t time, we have :

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

1: When v = 0, we have v = 2t – 2 = 0

$$\begin{array}{l}\Rightarrow t=1s\end{array}$$

2. Velocity is expressed in terms of time as :

$$\begin{array}{l}v=2t-2\end{array}$$

It is clear from the expression that velocity is negative for t<1 second, while positive for t>1. As such velocity changes its direction during motion.

3: Positions of the particle at successive seconds for first three seconds are :

$$\begin{array}{l}t=0;x=t^2-2t+1=0-0+1=1m\\ t=1;x=t^2-2t+1=1-2+1=0m\\ t=2;x=t^2-2t+1=4-4+1=1m\\ t=3;x=t^2-2t+1=9-6+1=4m\end{array}$$

4: Positions of the particle at t = 0 and t = 3 s are 1 m and 4 m from the origin.

Hence, displacement in first three seconds is 4 – 1 = 3 m

5: The particle moves from the start position, x = 1 m, in the negative direction for 1 second. At t = 1, the particle comes to rest. For the time interval from 1 to 3 seconds, the particle moves in the positive direction.