One Dimensional Kinematics: Motion Along a Straight Line

What do we mean by a "moving" body? If a body changes its position with time, it is said to be moving. But how can we say that the body is changing it's position? Well, it depends on the observer. Say, you are sitting on a bench, your eyes are working properly and a lousy cat is sitting near you. By seeing the cat, you can say that it is at rest, i.e., it is not moving. But if you imagine yourself on the moon, watching the same lousy cat with the help of a very good telescope, you say that the cat is moving (because of the Earth's motion with respect to moon). Remember, when you say cat is moving or not, it doesn't matter that the cat is "moving" its legs, we only focus on the change in POSITION of the cat.

So far, we have established that motion is a combined property of object under motion and the observer.There is no meaning of motion without the observer.

We are about to discuss the simplest type of motion, i.e., motion in a straight line.

Now, when a particle is restricted to move on a straight line, We generally choose the line on which particle is moving as X-axis and a suitable moment of time as t=0. Generally the point at which the particle is situated at t=0 is taken as origin. The position of the particle at the time t is given by coordinate x at that time.

The velocity of the particle is \(v=\frac{dx}{dt}\) The acceleration is \(a=\frac{dv}{dt} =\frac{d^2x}{dt^2} \)

A particle is moving on X-axis with it's x-coordinate given at time t by - \(x=(t^2 + t + 1)m\) Find the position, velocity and acceleration of the particle at \(t=0\) and\(t=1s\).

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Given that the x-coordinate of the particle varies with time t as - \(x=(t^2 + t + 1)m\) ...... (i)

So, It's velocity at time \(t\) will be - \(v=\frac{dx}{dt} = \frac{d(t^2 + t + 1)}{dt} = (2t+1)m/s\) ..... (ii)

And acceleration at time \(t\), \(a=\frac{dv}{dt} = \frac{d(2t+1)}{dt} = 2m/s\) which is not dependent on time, i.e., constant.

Put values of \(t\) in the equations (i) and (ii) to get the position and the velocity of the particle at any moment of time.

Now, we are going to study a special case of motion in a straight line, which is, with constant acceleration.

Suppose the acceleration of the particle is \(a\) which is constant. Let the velocity at \(t=0\) be \(u\), i.e., \(v(t=0) = u\). Thus,

\(a=\frac{dv}{dt}\) or \(dv=adt\)

Integrating from \(t=0\) to \(t=t\), \(\int _{ u }^{ v(t) }{ dv } =\quad \int _{ 0 }^{ t }{ a dt } \)

We get, \(v(t) - u = a(t-0)\) or \(\boxed{v(t)= u + at}\) ...... (i)

This equation can be written as \(\frac{dx}{dt}= u + at\) or \( dx = (u + at)dt\)

Integrating from \(t=0\) to \(t=t\),

We get, \( x(t) - x(t=0) = u(t-0) + \frac{a(t-0)^2}{2}\)

or , \( x(t) - x(t=0) = ut + \frac{at^2}{2}\)

If \(x(t=0) = 0\), then, \( \boxed{x(t) = ut + \frac{at^2}{2}}\) .... (ii)

\(NOTE\) - Sometimes, \(v(t)\) is simply written as \(v\) and \( x(t)\) as \( x\) when \(x(t=0) = 0\)

Try to eliminate variable \(t\) from (i) and (ii) to get , \(\boxed{v^2 - u^2 = 2ax}\)

Example Question 2

A particle starts from \(x=0\) with an initial velocity \(2.5m/s\) along the positive x direction and it accelerates uniformly at the rate \(0.5 m/s^2\).

(a) Find the distance travelled by it in the first two seconds.

(b) How much time does it take to reach the velocity \(7.5 m/s\)?

(c) How much distance will it cover in reaching the velocity \(7.5 m/s\)?

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Firstly, we should check whether the particle will turn back or not.

As the initial velocity of the particle, \(u\) is along the positive x direction and the acceleration is also in the same direction, the velocity keeps increasing in positive x direction and the particle never turns back.

We have, \( x(t) = ut + \frac{at^2}{2}\)

\( x(t=2 s) = (2.5 m/s)(2 s) + \frac{((0.5 m/s^2)(2 s)^2}{2}\)

\( x(t=2 s) = 5 m + 1 m = 6 m\)

So, distance covered = displacement ( as the particle never turns back) = \( x(t=2 s) - x(t=0) = 6 m - 0 m = \boxed{6m}\)

From \(v(t)= u + at\),

\(7.5 m/s = 2.5 m/s + (0.5 m/s^2)(t)\)

On solving, we will get, \(\boxed{t=10 s}\)

For (c) part, we have \(v^2 - u^2 = 2ax\)

Putting \(v= 7.5 m/s , u = 2.5 m/s\) and \(a = 0.5 m/s^2\) we get, \(\boxed{x=50 m}\)

Remember that \(x\) represents the position of the particle at time \(t\) and not (in general) the distance traveled by it in time \(0\) to \(t\).

Take, for example, if a particle starts from \(x=2 m\) at \(t=0 s\) and goes upto \(x=9 m\) at time \(t s\) , then the distance traveled by it in \(0 s\) to \(t s\) is \(9 m-2 m= 7 m\) but the position is given by \(x=9 m\) at time \(t\)

Also, more general formula is
\(x({ t }_{ 2 }) - x({ t }_{ 1 }) = u({ t }_{ 2 } - { t }_{ 1 }) + \frac{1}{2}a({ t }_{ 2 } - { t }_{ 1 })^2\)

This equation is useful when you know \(u\) and \(a\), and want to calculate the displacement of the particle in time \({ t }_{ 1 }\) to \({ t }_{ 2 }\).

Then, displacement =\(\Delta x\) = \(x({ t }_{ 2 }) - x({ t }_{ 1 }) = u({ t }_{ 2 } - { t }_{ 1 }) + \frac{1}{2}a({ t }_{ 2 } - { t }_{ 1 })^2\)

We live on the Earth, so you are quite familiar with the gravity near the earth surface. For example, when you throw a ball straight up (nearly), it goes up, slows down, stops for an instant, starts moving downwards, speed increases and falls back on the surface. (Then It bounces back up, and so on, but here we will only describe it's motion until the ball (or body)touches the surface for first time. We say that the ball is "freely falling". When we use the expression freely falling object, we do not necessarily refer to an object dropped from rest.

Free fall

A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion. Objects thrown upward or downward and those released from rest are all falling freely once they are released. Any freely falling object experiences an acceleration directed downward, regardless of its initial motion.

If we neglect air resistance and assume that the free-fall acceleration does not vary with altitude over short vertical distances, then the motion of a freely falling object moving vertically is equivalent to motion in one dimension under constant acceleration.

We usually denote the magnitude of the free-fall acceleration by the symbol \(g\). It is common to define “up” as the positive \(y\) direction and to use \(y\) as the position variable in the kinematic equations. At the Earth’s surface, the value of \(g\) is approximately \(9.80 m/s^2\). For making quick estimates, use \(g=10 m/s^2 \).

So, the equations that we've derived for motion in one dimension are valid for this case. The only modification that we need to make in these equations for freely falling objects is to note that the motion is in the vertical direction (the \(y\) direction) rather than in the horizontal \(x\) direction and that the acceleration is downward and has a magnitude of \(9.80 m/s^2\). Thus, \(a = -g = -9.8 m/s^2 \).

Example Question 3

A ball is thrown up at a speed of \(4 m/s\) from the ground. Find the maximum height reached by the ball. Take \(g=10 m/s^2 \).

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Take vertically upward direction as positive \(y\) direction, \(y=0\) at ground, we have \(u=4 m/s\), \(a=-g=-10 m/s^2 \).

When the ball reaches highest point, its velocity becomes zero. Using \(v^2 = u^2 + 2 ay\),

for finding maximum height, \(v=0\) \( 0 = 4^2 + 2(-10)y\)

\(y = 0.8 m \)

So, maximum height reached by the ball is \(\boxed{0.8 m}\)

Example Question 4

A stone is dropped from rest from the top of a tall building, as shown in the figure. After 3.00 s of free-fall, what is the displacement y of the stone? use \(g=9.8 m/s^2 \)

falling ball

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Here, \(u=0\) and \(a=-g=-9.8 m/s^2 \).

\(\Delta y = 0(3) + \frac{1}{2} (-9.8)(3)^2\)

\(\Delta y = \boxed{-44.1 m}\)