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.
\nDistance = velocity x time
\nIf acceleration is involved in the question, the equation becomes
\nDistance = v0 x t + 0.5 a t^2
\nWhere v0 is original velocity.
\nWith these two equations, you can solve almost every question.<\/p>\n
Example<\/strong><\/p>\n If a car goes 3m\/s and it accelerates 2m\/s^2 for 5 seconds. What is the distance the car went? D = 3 x 5 + 0.5 x 2 x 5^2 Like this you can simply apply the values in the equation.<\/p>\n Example 2<\/strong><\/p>\n 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? <\/p>\n Like this you can apply the equation and solve the question.<\/p>\n <\/p>\n Example 3:<\/strong> An object falling.<\/p>\n 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? Also, a = 9.81m\/s^2, d=10 D = 0.5at^2 10 = 0.5*9.81*t^2 <\/p>\n Example 4:<\/strong><\/p>\n 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?<\/p>\n First, identify the values. We know that v0 = 0 since it\u2019s falling. Now, substitute the value in the equation
\nFirst, identify the values. a = 2,v0=3, and t=5
\nNow, substitute the value in the equation<\/p>\n
\n= 15 + 25
\n= 40<\/p>\n
\nFirst, identify the values. a = 4,d=40, and t=4
\nNow, substitute the value in the equation
\n40 = v0 x 4 + 0.5 x 4 x 4^2
\n40=4v0 + 32
\n8 = 4v0
\nv0 = 2
\n=> X = 2m\/s<\/p>\n
\nFirst, identify the values. We know that v0 = 0 since it\u2019s falling.<\/p>\n
\nNow, substitute the value in the equation<\/p>\n
\nWe know D = 10m, and a = 9.81m\/s^2<\/p>\n
\nt^2 = 5.095
\nt = 2.247seconds<\/p>\n
\nAlso, a = 4m\/s^2, d=20<\/p>\n
\nD = 0.5a*t^2
\n20=0.5*4*t^2
\n20=2t^2
\nt^2=10
\nt= 3.16s<\/p>\n