A brief note on Acceleration and the Equations of Motion . Try each question at the end before you open its answer.
The big ideas Acceleration is the rate of change of velocity: a = v − u t a = \dfrac{v - u}{t} a = t v − u , in m/s 2 \text{m/s}^2 m/s 2 . Speeding up or slowing down? Same signs of v v v and a a a : speeding up. Opposite signs: slowing down. Velocity-time graph: the slope is the acceleration; the area under the line is the displacement (negative below the axis). For a steady acceleration, three equations answer almost every question. Equation It leaves out v = u + a t v = u + at v = u + a t s s s s = u t + 1 2 a t 2 s = ut + \tfrac12 at^2 s = u t + 2 1 a t 2 v v v v 2 = u 2 + 2 a s v^2 = u^2 + 2as v 2 = u 2 + 2 a s t t t s = u + v 2 t s = \dfrac{u + v}{2}\,t s = 2 u + v t a a a
And two results built from them:
d stop = v t r + v 2 2 b \displaystyle d_{\text{stop}} = v\,t_r + \frac{v^2}{2b} d stop = v t r + 2 b v 2 s n = a 2 ( 2 n − 1 ) \displaystyle s_n = \frac{a}{2}(2n - 1) s n = 2 a ( 2 n − 1 ) The first is the stopping distance (thinking plus braking); the second, the distance in the n n n th second from rest.
A worked example A train comes into a station at 22 m/s 22\ \text{m/s} 22 m/s and brakes steadily, stopping in 242 m 242\ \text{m} 242 m .
Deceleration. 0 = ( 22 m/s ) 2 + 2 a × 242 m 0 = (22\ \text{m/s})^2 + 2a \times 242\ \text{m} 0 = ( 22 m/s ) 2 + 2 a × 242 m , so a = − 484 m 2 / s 2 484 m = − 1.0 m/s 2 a = -\dfrac{484\ \text{m}^2/\text{s}^2}{484\ \text{m}} = -1.0\ \text{m/s}^2 a = − 484 m 484 m 2 / s 2 = − 1.0 m/s 2 . Time. 0 = 22 m/s − 1.0 m/s 2 × t 0 = 22\ \text{m/s} - 1.0\ \text{m/s}^2 \times t 0 = 22 m/s − 1.0 m/s 2 × t , so t = 22 s t = 22\ \text{s} t = 22 s . Check. Average velocity 11 m/s 11\ \text{m/s} 11 m/s for 22 s 22\ \text{s} 22 s gives 242 m 242\ \text{m} 242 m . Practice with new numbers
A program to try Give it u u u , a a a and s s s ; it finds v v v and t t t from the equations.
Predict first: if the platform were only 200 m long, would the train stop on it? Change s and run it.
What does the program print, and why does it need max(v_squared, 0)?
Mistakes to avoid "Negative acceleration means slowing down." Only when the velocity is positive. A car reversing faster has a negative acceleration too. Signs. Choose a positive direction once, and give u u u , v v v , a a a and s s s their signs. Forgetting the 1 2 \tfrac12 2 1 in 1 2 a t 2 \tfrac12 at^2 2 1 a t 2 , or the 2 in 2 a s 2as 2 a s . Braking distance and speed. Double the speed, four times the braking distance. "In the 5th second" is one second, not five. Check yourself A cyclist slows from 9 m/s 9\ \text{m/s} 9 m/s to 3 m/s 3\ \text{m/s} 3 m/s in 4 s 4\ \text{s} 4 s . Find her acceleration. Stuck? Show a hint Show the answerFrom rest, a cart accelerates at 0.8 m/s 2 0.8\ \text{m/s}^2 0.8 m/s 2 for 10 s 10\ \text{s} 10 s . How far does it go? Stuck? Show a hint Show the answerA velocity-time graph rises in a straight line from 0 m/s 0\ \text{m/s} 0 m/s to 14 m/s 14\ \text{m/s} 14 m/s in 7 s 7\ \text{s} 7 s . What are the acceleration and the distance? Stuck? Show a hint Show the answerWith the same brakes, a car's speed goes from 10 m/s 10\ \text{m/s} 10 m/s to 15 m/s 15\ \text{m/s} 15 m/s . How many times longer is its braking distance? Stuck? Show a hint Show the answerFrom rest at 2 m/s 2 2\ \text{m/s}^2 2 m/s 2 , how far does a body go in the 3 3 3 rd second? Stuck? Show a hint Show the answer