Motion in a Straight Line & Motion in a Plane

Motion in a Straight Line & Motion in a Plane

Motion in a Straight Line and Motion in a Plane form the launching pads in Physics. Rectilinear motion is effectively the first lesson in Kinematics. In Kinematics, we study ways to describe motion without going into the causes of motion. Let’s recap the fundamentals first. Click to study Quadratic Equations.


For objects in uniformly accelerated rectilinear motion, the five quantities, displacement \( s \), time taken \( t \), initial velocity \( u \), final velocity \( v \) and acceleration \( a \) are related by a set of simple equations called kinematic equations of motion given below.

\[ v = u + at \] \[ s = ut + \frac{1}{2}at^2 \] \[ v^2 = u^2 + 2as \] \[ s = \left( \frac{u + v}{2} \right) \] \[ S_n = u - \frac{a}{2}(2n - 1) \]

where \( u = \) initial velocity, \( v = \) final velocity, \( a = \) acceleration, \( s = \) displacement and \( S_n = \) displacement in the \( n^{th} \) second, and the position of the object at time \( t = 0 \) is \( 0 \). If the particle starts at \( s = s_0 \), \( s \) in above equations is replaced by \( (s - s_0) \). The modern practice is to use \( v_0 \) for initial velocity and \( x \) for displacement. Earlier, \( f \) was used for acceleration. As you wish!


#1. An object is said to be in motion if its position changes with time.

#2. Dimensions of Motion: The three coordinates \( x, y, z \) specify the position of an object in space. If only one of the coordinates out of the three changes with respect to time, it is 1-D motion, e.g an object falling freely under gravity. Motion in 1 dimension is motion along a straight line, the subject of our present discussion. If two of these change, it is 2-D motion, e.g circular motion, and when all three coordinates change, it is 3-D motion, e.g random motion of gas molecule. In general, motion in space is three-dimensional; the others are more of theoretical constructs.

#3. Displacement is the change in position : \( \newcommand{\changeIn}{\operatorname{\Delta}}\changeIn(x) = x_2 - x_1 \)

#4. Distance traversed is greater than or equal to displacement. Distance is scalar quantity whereas displacement is vector.

#5. Average speed is greater than or equal to average velocity. Speed is scalar quantity whereas velocity is vector. Dimensional formula of speed and velocity: \( [\text{M }^0\text{ T }^{-1}] \)

#6. Average velocity of an object depends only on its initial and final location and time, and truly not on the motion of the object in between. It as such is in general not a useful parameter. A more useful quantity is the instantaneous velocity of an object at a given instant. The instantaneous velocity is the value that the average velocity approaches as the time interval over which it is measured approaches zero. Said otherwise, instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval \( \newcommand{\changeIn}{\operatorname{\Delta}}\changeIn(t) \) becomes infinitesimally small: \[ \newcommand{\changeIn}{\operatorname{\Delta}} v = \lim_{\changeIn(t) \rightarrow 0} {\overline{v} = \lim_{\changeIn(t) \rightarrow 0} {\frac{\changeIn(x)}{\changeIn(t)}}} \]

#7. Velocity is defined in terms of change of position of the object over time. The change of velocity over time is called acceleration. Average acceleration: \( \newcommand{\changeIn}{\operatorname{\Delta}}\overline{a} = \frac{\changeIn(v)}{\changeIn(t)} \)

#8. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval \( \newcommand{\changeIn}{\operatorname{\Delta}}\changeIn(t) \) goes to zero : \[ \newcommand{\changeIn}{\operatorname{\Delta}} a = \lim_{\changeIn(t) \rightarrow 0} {\overline{a} = \lim_{\changeIn(t) \rightarrow 0} {\frac{\changeIn(v)}{\changeIn(t)} = \frac{dv}{dt}}} \]

#9. Constant acceleration: Objects falling freely under the influence of gravity (falling in vacuum) are one example of objects moving with constant acceleration. A constant acceleration means that the acceleration does not depend on time: \( a = \frac{dv}{dt} = k \)

#10. The sign of acceleration does not tell us whether the particle’s speed is increasing or decreasing. The sign of acceleration depends on the choice of the positive direction of the axis. For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed. For a particle thrown upward, the same negative acceleration (of gravity) results in decrease in speed.

#11. The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have non-zero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity.



A vehicle travels half the distance \( L \) with speed \( V_1 \) and the other half with speed \( V_2 \), then its average speed is:

  1. \( \frac{V_1 + V_2}{2} \)
  2. \( \frac{V_1 + V_2}{V_1V_2} \)
  3. \( \frac{V_1V_2}{V_1 + V_2} \)
  4. \( \frac{2V_1V_2}{V_1 + V_2} \)


A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?

  1. \( x < 0, v < 0, a > 0 \)
  2. \( x > 0, v < 0, a < 0 \)
  3. \( x > 0, v < 0, a > 0 \)
  4. \( x > 0, v > 0, a < 0 \)


The displacement of a particle is given by \( x = (t – 2)^2 \) where \( x \) is in metres and \( t \) in seconds. Find the distance covered by the particle in first 4 seconds in metres.


A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of 18 km/h while the other has the speed of 27km/h. The bird starts moving from first car towards the other and is moving with the speed of 36km/h and when the two cars were separated by 36 km. What is the total distance covered by the bird? What is the total displacement of the bird?


It is a common observation that rain clouds can be at about a kilometre altitude above the ground. If a rain drop falls from such a height freely under gravity, what will be its speed in km/h? ( g = 10m/s²)


At a metro station, a girl walks up a stationary escalator in time t1. If she remains stationary on the escalator, then the escalator takes her up in time t2. The time taken by her to walk up on the moving escalator will be

  1. (t1 + t2)/2
  2. t1.t2/(t2–t1)
  3. t1.t2/(t2+t1)
  4. t1–t2

Relative Motion

Let us start with a very simple poser. Let’s do it fast and without pen and paper.


A thief is running away on a straight road in a four-wheeler with uniform velocity of 9ms-1. A policeman chases him on a bike with un-accelerated constant speed of 10ms-1. After what time will the policeman catch the thief if he is 100m behind him at the time of reporting?

Now, some discussions:

We need a frame of reference to locate an object. We need a frame of reference to plot motion of an object over time. This frame of reference might be the observer. Thus, motion is the combined property of the object and the observer. Without frame of reference or the observer both rest and motion are meaningless. So, rest and motion are relative terms. Now, the point of reference or the observer might also be in motion relative to the earth. There comes to fore the concept of relative motion.

The fundamental equations of kinematics are modified in the case of relative motion. Suppose two objects A and B are moving along the same line with velocities of magnitude VA and VB relative to the earth. We have to find the velocity of B relative to A. We denote this Vrel as VBA. When the two bodies are moving in the direction this is given by VBA = VB – VA. When the result is +ve, the direction of VBA is that of B, when –ve, opposite to B. Objects A and B might be moving towards or away from each-other. In that case, VBA = VB + VA. To get our answers correct, we must put the known values with correct signs. This was explaining one-dimensional relative motion. Similarly, we can proceed to relative motion in 2D.

To put the above in plain English, the relative velocity of two bodies themselves in motion is equal to the difference if they are moving in the same direction and is the sum when they are moving in the opposite directions.

Relative Motion in Point Object and Plane Mirror

There can be two cases. First, the simple one: the object in motion and the mirror can be parallel to each-other. Then, velocity of image will be equal to the velocity of the object and in the same direction. Often that is not the case. When the mirror is perpendicular, the velocity of image with respect to mirror will be equal and opposite of the velocity of the object with respect to the mirror.

Right now, we move on to the questions part.


Two parallel rail tracks run north-south. Train A moves north with a speed of 54 km/h, and train B moves south with a speed of 90 km/h. What is the (a) velocity of B with respect to A, (b) velocity of ground with respect to B, and (c) velocity of a monkey running on the roof of the train A against its motion (with a velocity of 18 km/h with respect to the train A) as observed by a man standing on the ground?


A ball is dropped from a building of height 45 m. Simultaneously, another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.


Rain is falling vertically with a speed of 35 m/s. Winds start blowing after sometime with a speed of 12 m/s in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella?


Re-do #4 with slight but significant variation. Rain is falling vertically with a speed of 35 m/s. A woman rides a bicycle with a speed of 12 m/s in east to west direction. What is the direction in which she should hold her umbrella?

Projectile Motion

An intuitive poser, first.


Which of the following is not an example of projectile? (A) A hydrogen-filled balloon moving up in air, (B) A bullet fired from a gun, (C) A ball thrown vertically upwards, (D) Helicopter shot by MSD


Projectile motion is motion in a plane. Motion in a plane is 2-dimensional or 2D motion. It can be understood as vector sum of two independent 1D motions along two mutually perpendicular directions. There is nothing common in between the two except of course time. So, the particle is moving in x-y plane. It has two independent motions: one in x-axis, another in y.

In one dimensional motion, the velocity and the acceleration of an object are always along the same straight line (either in the same direction or in the opposite direction). However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° between them. The essential formulae for x-axis will be: \[ v_x = u_x + a_xt \]\[ s_x = u_xt + \frac{1}{2}at^2 \]\[ (v_x)^2 = (u_x)^2 + 2a_xx \] and the same for y-axis will be
\[ v_y = u_y + a_yt \]\[ s_y = u_yt + \frac{1}{2}at^2 \]\[ (v_y)^2 = (u_y)^2 + 2a_yx \]

The motion of an object, after being thrown, in flight through the atmosphere under the effect of gravity alone is called projectile motion. Air resistance is neglected.

PM is a 2D motion. The horizontal component is an unchanged uniform motion without any acceleration. The vertical component is uniformly accelerated motion. The acceleration can be positive or negative. PM can be of three types: oblique, horizontal and on inclined plane.

The position and velocity of a projectile at time \( u \) are given by: \[ x = (u\cos\theta) t \] \[ y = (u\sin\theta) t – \frac{1}{2}gt^2 \] \[ v_x = u_x = u\cos\theta \] \[ v_y = u\sin\theta – gt \] The path of a projectile is parabolic and given by \[ y = (\tan\theta)x – \frac{g}{2}\left( \frac{x}{u\cos\theta} \right)^2
\] Below are given formulae for range, height and time of flight: \[ R = \frac{u^2\sin2\theta}{g} \] \[ H = \frac{u^2\sin^2\theta}{2g} \] \[ T = \frac{2u\sin\theta}{g} \]
Range maximum for \( \theta = 45^0 \).
Height maximum for \( \theta = 90^0 \).
Angular Momentum at peak \(L = mvH\).



The trajectory of a projectile is given by \( y = \sqrt{3x} – \frac{gx^2}{2} \). What is the angle of projection?

Hint: Compare the equation with the standard equation
\[ y = x\tan \theta – \frac{gx^2}{2u^2\ cos^2\theta} \]


From the edge of a cliff 490 m above the ground, a hiker throws a stone horizontally with an initial speed of 15 m s-1. Find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2).


A cricket ball is thrown at a speed of 28 m/s in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level.


The horizontal range of a projectile fired at an angle of 15° is 50 m. What will be its range if it is fired with the same speed at an angle of 45°?


A ball is thrown from a roof top at an angle of 45° above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have
(a) greatest speed.
(b) smallest speed.
(c) greatest acceleration?


If a football is kicked into the air vertically upwards, what is its (a) acceleration, and (b) velocity at the highest point?


In dealing with projectile motion, we ignore effect of air resistance. This gives trajectory as a parabola. What would the trajectory look like if air resistance is included? Sketch and explain.


Hover over here.

Best of luck!
Straight Line
1. Opt. 4. \( \frac{2V_1V_2}{V_1 + V_2} \)
2. Opt. 1. \( x < 0, v < 0, a > 0 \)
3. Ans: 8 m
4. Ans: 28.8 km
5. Ans: 510 km/h
6. Option 3. \( \left( \frac{t_1 \times t_2}{t_1 + t_2} \right) \)
Relative Motion
1. 100 seconds
2. (a) 40 m/s To South (b) 25 m/s To North (c) 10 m/s To North
3. Remains constant @ 40m/s
4. At about 19° with the vertical towards East.
5. Same as #4, only towards West.
Projectile Motion
1. Opt. A. Balloon
2. Ans: \( \theta = 60^0 \)
3. 99 m/s
4. (a) 10 m (b) 2.9 s (c) 69 m
5. 100 m
6. (a) Just before it hits the ground.(b) At the highest point reached. (c) a = g = constant.
7. Acceleration: g. Velocity: zero.
8. Fall is steeper than rise.

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