IB — physicsFoundational3 min read

Uniformly Accelerated Motion

SUVAT equations; free fall under gravity; solving kinematics problems

Tutor AI Team
Tutor AI Team

# Uniformly Accelerated Motion — IB Physics

When acceleration is constant, we can use the SUVAT equations to relate displacement, velocity, acceleration, and time. This applies to many real situations, including free fall.

1. SUVAT Equations

v=u+atv = u + at s=ut+12at2s = ut + \frac{1}{2}at^2 s=(u+v)2ts = \frac{(u + v)}{2}t v2=u2+2asv^2 = u^2 + 2as s=vt−12at2s = vt - \frac{1}{2}at^2

Symbol Meaning
ss Displacement
uu Initial velocity
vv Final velocity
aa Acceleration
tt Time

2. Free Fall

Near Earth's surface: a=g=9.81a = g = 9.81 m/s² (downward).

Sign convention: choose one direction as positive (usually up or down) and be consistent.

Dropping: u=0u = 0, a=ga = g downward. Throwing up: u>0u > 0 upward, a=ga = g downward. At maximum height, v=0v = 0.

Worked Example: Car Braking

Problem

A car at 25 m/s brakes at −5 m/s². Find the stopping distance.

Known: u=25u = 25, v=0v = 0, a=−5a = -5. Find ss. v2=u2+2asv^2 = u^2 + 2as 0=625+2(−5)s0 = 625 + 2(-5)s s=62.5s = 62.5 m

Solution

Worked Example: Free Fall

Problem

A ball is dropped from 45 m. Find the time to hit the ground and impact speed.

s=ut+12gt2s = ut + \frac{1}{2}gt^2: 45=0+12(9.81)t245 = 0 + \frac{1}{2}(9.81)t^2 t=90/9.81=3.03t = \sqrt{90/9.81} = 3.03 s v=u+gt=0+9.81(3.03)=29.7v = u + gt = 0 + 9.81(3.03) = 29.7 m/s

Solution

Worked Example: Throw Up

Problem

A ball is thrown up at 20 m/s. Find maximum height and total time of flight.

At top: v=0v = 0. v2=u2−2gsv^2 = u^2 - 2gs (taking up as +) 0=400−2(9.81)s0 = 400 - 2(9.81)s → s=20.4s = 20.4 m

Time up: v=u−gtv = u - gt → t=20/9.81=2.04t = 20/9.81 = 2.04 s Total time = 2×2.04=4.082 \times 2.04 = 4.08 s (symmetric)

Solution

4. IB Exam Tips

  • Always list known values before choosing an equation
  • Choose the equation that includes the three known values and the unknown
  • Watch sign conventions carefully
  • In IB, g=9.81g = 9.81 m/s² (or sometimes 9.8 or 10 for estimation)

5. Practice Questions

    1. A car accelerates from rest at 3 m/s² for 8 s. Calculate (a) final velocity, (b) distance covered. (4 marks)
    1. A stone is dropped from a bridge and hits the water 2.5 s later. Find the height of the bridge. (2 marks)
    1. A rocket launches vertically with acceleration 15 m/s² for 10 s, then the engines cut. Find maximum height. (5 marks)

    Answers

    1. (a) v=0+3(8)=24v = 0 + 3(8) = 24 m/s. (b) s=0+12(3)(64)=96s = 0 + \frac{1}{2}(3)(64) = 96 m.
    1. s=12(9.81)(6.25)=30.7s = \frac{1}{2}(9.81)(6.25) = 30.7 m.

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Summary

  • SUVAT equations valid for constant acceleration
  • Free fall: a=g=9.81a = g = 9.81 m/s² downward
  • Carefully choose sign convention and list knowns before solving
Tags:
#ib#physics#mechanics#suvat#acceleration#free-fall#kinematics

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