Section 1.4: Constant Acceleration Equations
When acceleration is constant, motion can be described with a small set of kinematic equations that connect displacement, velocity, acceleration and time. These are essential for solving many 1-D motion problems.
- Motion is along a straight line (1D).
- Acceleration \(a\) is constant during the interval considered.
Key Equations
For constant acceleration \(a\):
- \(v = v_0 + a t\) — velocity after time \(t\).
- \(x = x_0 + v_0 t + \dfrac{1}{2} a t^2\) — position after time \(t\).
- \(v^2 = v_0^2 + 2 a (x - x_0)\) — relation without time.
- \( \bar{v} = \dfrac{v_0 + v}{2} \) — average velocity for constant acceleration.
Example 1
A car has initial speed \(v_0 = 5\ \mathrm{m/s}\) and accelerates at \(3\ \mathrm{m/s^2}\). Find the speed after \(4\) s and the distance traveled in that time.
\( v = v_0 + a t = 5 + 3(4) = 17\ \mathrm{m/s} \).
\( x - x_0 = v_0 t + \tfrac{1}{2} a t^2 = 5(4) + \tfrac{1}{2}(3)(16) = 20 + 24 = 44\ \mathrm{m} \).
Example 2 (Using \(v^2\) relation)
A cyclist increasing speed from \(2\ \mathrm{m/s}\) to \(10\ \mathrm{m/s}\) with constant acceleration covers 49 m. Find the acceleration.
Use \(v^2 = v_0^2 + 2 a (x - x_0)\):
\(10^2 = 2^2 + 2 a (49)\) → \(100 = 4 + 98a\) → \(98a = 96\) → \(a = \dfrac{96}{98} \approx 0.9796\ \mathrm{m/s^2}.\)
Practice Problems
- A car starts from rest and accelerates uniformly at \(2.5\ \mathrm{m/s^2}\). How far does it travel in 6 s?
- A stone thrown upward with initial speed \(14\ \mathrm{m/s}\). Using \(g=9.8\ \mathrm{m/s^2}\) downward, find maximum height (use \(v^2\) relation).
- A train slows from \(25\ \mathrm{m/s}\) to \(10\ \mathrm{m/s}\) over 60 s. Find its acceleration and distance covered in that time.
- An object moves with \(x(t)=2+4t+3t^2\). Verify which constant acceleration equation it satisfies and find \(a\), \(v_0\), and displacement from \(t=0\) to \(t=3\) s.