Distance, Velocity, and Acceleration
There is a relationship between distance, velocity and acceleration. When you understand how they interact, and their equations; they are much easier to grasp. The relationship between distance and velocity is proportional. Distance = velocity x time If acceleration is involved in the question, the equation becomes Distance = v0 x t + 0.5 a t^2 Where v0 is original velocity. With these two equations, you can solve almost every question. Example If a car goes 3m/s and it accelerates 2m/s^2 for 5 seconds. What is the distance the car went? First, identify the values. a = 2,v0=3, and t=5 Now, substitute the value in the equation D = 3 x 5 + 0.5 x 2 x 5^2 = 15 + 25 = 40 Like this you can simply apply the values in the equation. Example 2 A car was going X m/s. It accelerates 4m/s^2 for 4 seconds. The total distance it went was 40m. What is X? First, identify the values. a = 4,d=40, and t=4 Now, substitute the value in the equation 40 = v0 x 4 + 0.5 x 4 x 4^2 40=4v0 + 32 8 = 4v0 v0 = 2 => X = 2m/s Like this you can apply the equation and solve the question. Example 3: An object falling. Newton dropped an apple from 10m high. How long would it take for it to touch the ground if the acceleration is 9.81m/s^2? First, identify the values. We know that v0 = 0 since it’s falling. Also, a = 9.81m/s^2, d=10 Now, substitute the value in the equation D = 0.5at^2 We know D = 10m, and a = 9.81m/s^2 10 = 0.5*9.81*t^2 t^2 = 5.095 t = 2.247seconds Example 4: Galileo dropped a feather from 20m high. The acceleration it falls is 4m/s^2. How long would it take for the feather to touch the ground? First, identify the values. We know that v0 = 0 since it’s falling. Also, a = 4m/s^2, d=20 Now, substitute the value in the equation D = 0.5a*t^2 20=0.5*4*t^2 20=2t^2 t^2=10 t= 3.16s Tutors for Physics are available if you need more help. This article was written for you by Edmond, one of the tutors with SchoolTutoring Academy.