Equations of Motion and Gravity: PECTAA Class 11 Physics, Punjab Board

A car speeds up. A ball falls. A runner leaves the blocks. Each is motion in a straight line with steady acceleration. Three short equations cover all of them. Learn what each symbol means, and which equation to pick, and the numericals take two minutes.

Chapter 2 (Force and Motion) of the PECTAA Physics 11 book gives the equations of motion in section 2.4 and motion under gravity in section 2.5. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Take a body moving along a straight line with uniform acceleration a. Its initial velocity is v_i, its final velocity v_f after time t, and its displacement is S.

First equation. On the velocity-time graph the line AB rises from v_i to v_f. The final velocity is DB = DC + CB. Since OA = DC = v_i, we get v_f = v_i + CB. The slope of AB is the acceleration, a = CB ÷ t, so CB = at. Hence v_f = v_i + at.

Second equation. Displacement = average velocity × time, so S = ½(v_i + v_f)t. Put v_f = v_i + at to get S = v_i t + ½at². The graph gives the same result: S is the area of the rectangle (v_i × t) plus the area of the triangle (½ × t × at).

Third equation. From 2S = (v_i + v_f)t and v_f = v_i + at, eliminating t gives 2aS = v_f² − v_i².

Motion under gravity. A body falling under gravity alone is the most familiar example of uniformly accelerated straight-line motion. According to Galileo, all bodies fall in a vacuum with the same acceleration due to gravity, g. The experimental value is 9.8 m s⁻² in SI units. So bodies dropped from the same height strike the ground with the same velocity. The book puts a = g in the three equations:

• v_f = v_i + gt
• h = v_i t + ½gt²
• 2gh = v_f² − v_i²

Sign of g. The book takes g positive for a falling body (initial velocity zero) and negative for a body projected vertically upward.

Key points

• Pick the equation by the missing quantity. Write what is given and what is asked. Then choose the row in the table that has no unknown extra.
• Uniform acceleration only. These equations are for constant acceleration along a straight line.
• Convert units first. Change km h⁻¹ to m s⁻¹ before using the equations. Divide by 3.6, or multiply by 1000 and divide by 3600.
• Starting from rest means v_i = 0.
• Graph facts. On a velocity-time graph, slope is acceleration and area under the line is displacement.

Worked examples

Example 1 (book Example 2.2). A car at 10 m s⁻¹ accelerates uniformly at 2 m s⁻². Find its velocity after 5 s.

1. v_i = 10 m s⁻¹, a = 2 m s⁻², t = 5 s.
2. v_f = v_i + at = 10 + 2 × 5 = 20 m s⁻¹.

Example 2 (book Example 2.3). A car with initial velocity 15 m s⁻¹ accelerates at 2 m s⁻² for 4 s. Find the displacement.

1. S = v_i t + ½at² = 15 × 4 + ½ × 2 × 4².
2. S = 60 + 16 = 76 m.

Example 3 (book Example 2.4). A car starts from rest and reaches 300 km h⁻¹ after covering 0.45 km. Find the constant acceleration.

1. v_i = 0. v_f = 300 × 1000 ÷ 3600 = 83.33 m s⁻¹. S = 0.45 × 1000 = 450 m.
2. Use 2aS = v_f² − v_i²: 2 × a × 450 = (83.33)² − 0.
3. a = 6943.9 ÷ 900 = 7.72 m s⁻².

Example 4 (book example on motion under gravity). An iron ball is dropped from a tower and reaches the ground in 3.34 s. Find (a) the velocity on striking the ground and (b) the height of the tower.

1. v_i = 0, a = g = 9.8 m s⁻², t = 3.34 s.
2. (a) v_f = gt = 9.8 × 3.34 = 32.7 m s⁻¹.
3. (b) 2gh = v_f²: h = (32.7)² ÷ (2 × 9.8) = 1069.29 ÷ 19.6 = 54.56 m (the book rounds v to 32.7 first).
4. The mass of the ball (1 kg) is not needed.

Common mistakes in the exam

1. Using km h⁻¹ with metres and seconds. Convert first.
2. Using the equations for changing acceleration. They need constant a.
3. Forgetting v_i = 0 for "starts from rest" or "dropped".
4. Mixing up the sign of g. Follow the book: positive for falling from rest, negative for upward projection.
5. Using the mass in a falling-body question. All bodies fall with the same g.
6. Dropping units. Write m, m s⁻¹ and m s⁻² in the answer.

Exam technique

• Write the data list first: v_i, v_f, a, t, S. Mark the unknown with a question mark.
• Name the equation you use in words, as the book does ("Using first equation of motion").
• For derivation questions, draw the velocity-time graph and label O, A, B, C, D.
• Check your board's notice for the sections in your paper.

Practice questions

1. A body starts from rest and accelerates at 3 m s⁻² for 6 s. Find its final velocity. Answer: v_f = 0 + 3 × 6 = 18 m s⁻¹.
2. Find the displacement of the body in question 1. Answer: S = 0 + ½ × 3 × 36 = 54 m.
3. A car at 20 m s⁻¹ reaches 30 m s⁻¹ over 125 m. Find a. Answer: a = (900 − 400) ÷ (2 × 125) = 2 m s⁻².
4. A stone is dropped and hits the ground after 2 s. Find its velocity at impact (g = 9.8 m s⁻²). Answer: v_f = 9.8 × 2 = 19.6 m s⁻¹.
5. What does the area under a velocity-time graph give? Answer: The displacement.

Ready to practise this? Continue in 1st Year Physics.