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.