Motion in a Straight Line

In this chapter: position, path length and displacement, average and instantaneous velocity, speed, acceleration, reading x-t and v-t graphs, the kinematic equations for uniform acceleration, free fall, stopping distance, reaction time and relative velocity in one dimension.

Position, displacement and path length

To describe motion we need a reference point (the origin), a line with a chosen positive direction, and a clock. Together these make a frame of reference. In this chapter the object is treated as a point object, which is fine whenever its size is small compared with the distance it moves.

• Displacement is the change in position, Δx = x₂ − x₁. It has a sign and can be zero even when the object has moved.
• Path length (distance) is the total length of the path actually covered. It is always positive and never less than the magnitude of displacement.

If a car goes from x = 0 to x = +360 m and comes back to x = +240 m, its displacement is +240 m but the path length is 360 + 120 = 480 m.

Velocity and speed

Instantaneous velocity is the limit of the average velocity as the time interval shrinks to zero:

Instantaneous speed is the magnitude of instantaneous velocity. Over a finite interval this is not true for averages: a runner who finishes one lap of a track has zero average velocity but a large average speed.

On an x-t graph, a straight line means uniform velocity, a curve bending upward means velocity increasing with time, and a curve bending downward means velocity decreasing. A horizontal line means the object is at rest.

Acceleration

The sign of acceleration alone does not tell you whether the object is speeding up. Look at the signs of v and a together. If they have the same sign, the speed increases; if they have opposite signs, the speed decreases. A car moving in the negative direction and braking has negative v and positive a. Students often call every negative acceleration "retardation", which is wrong.

Also note that an object can have zero velocity and non-zero acceleration at the same instant, for example a ball at the top of its flight.

What the graphs tell you

For distance, add the magnitudes of the areas above and below the t-axis; for displacement, add them with their signs.

Kinematic equations for uniform acceleration

When a is constant, these relations connect initial velocity u, final velocity v, displacement x and time t:

The v-t graph gives a clean derivation. The slope of the line is a, so v = u + at. The area under it is a rectangle (ut) plus a triangle, ½(v − u)t = ½at², which gives the second equation. Eliminating t between the first two gives the third.

These equations hold only for constant acceleration. If a changes with time, go back to v = dx/dt and a = dv/dt and integrate.

Free fall and vertical motion

Neglecting air resistance, every body near the Earth's surface falls with the same acceleration g = 9.8 m s⁻², directed downward. Choose one direction as positive and stick to it. If upward is positive, then a = −g throughout the flight, both going up and coming down.

• For a body dropped from rest: v = gt, distance fallen = ½gt², v² = 2gh.
• For a body thrown up with speed u: maximum height = u²/2g, time to reach the top = u/g, and it returns to the same level with the same speed after 2u/g.
• Galileo's law of odd numbers: for a body starting from rest with uniform acceleration, the distances covered in successive equal time intervals are in the ratio 1 : 3 : 5 : 7 : ...

Worked example: A ball is thrown vertically upward with a speed of 20 m s⁻¹ from the top of a building 25.0 m high. Take g = 10 m s⁻². How high above the ground does it rise, and when does it hit the ground?
Solution: Take upward as positive, origin at the top of the building. At the highest point v = 0, so 0 = 20² − 2(10)y, giving y = 20 m. The maximum height above the ground is 25 + 20 = 45 m. When it hits the ground, y = −25 m: −25 = 20t − 5t², or t² − 4t − 5 = 0, so (t − 5)(t + 1) = 0 and t = 5 s (the negative root has no meaning here).

Stopping distance and reaction time

A vehicle moving at speed v₀ that brakes with uniform deceleration a stops in a distance d_s = v₀²/2a. Stopping distance goes as the square of the speed, so doubling the speed makes it four times longer.

Worked example: A car moving at 72 km h⁻¹ brakes with a uniform deceleration of 4 m s⁻². Find the stopping distance and time.
Solution: 72 km h⁻¹ = 72 × 5/18 = 20 m s⁻¹. d_s = 20²/(2 × 4) = 50 m. Time = 20/4 = 5 s.

Reaction time is the time a person takes to observe, think and act. You can measure it by dropping a ruler between a friend's fingers: if they catch it after it falls a distance d, then t_r = √2d/g. For d = 21.0 cm, t_r = √(2 × 0.21/9.8) ≈ 0.21 s.

Relative velocity in one dimension

If objects A and B move along the same line with velocities v_A and v_B, the velocity of B relative to A is

If v_A = v_B, their separation stays constant and the x-t graphs are parallel lines. Using relative velocity turns many two-body chase problems into a single-body problem with one equation.

Common mistakes: (1) Taking g as positive going up and negative coming down in the same problem; with upward positive, a = −g throughout. (2) Using the kinematic equations when the acceleration is not constant. (3) Reporting displacement when distance is asked, especially when the velocity changes sign during the interval. (4) Averaging speeds directly: for equal distances at v₁ and v₂, the average speed is 2v₁v₂/(v₁ + v₂), not (v₁ + v₂)/2. (5) Forgetting to convert km h⁻¹ to m s⁻¹ (multiply by 5/18).

JEE and NEET focus

• Reading slopes and areas of x-t, v-t and a-t graphs, including areas below the axis.
• Vertical projection problems: maximum height, time of flight, a body thrown from a tower.
• Distance in the n^th second and Galileo's ratio 1 : 3 : 5.
• Calculus-based problems where x or v is given as a function of t (JEE asks these often), using a = v dv/dx where needed.
• Average speed over equal distances and equal times, and relative velocity for trains and cars.

Practice questions