Enter exactly 3 values below to solve for the other 2 values.
Define positive to be right or upwards, negative to be left or downwards.
Scroll down after calculation to view solution and explanations.
Initial Velocity u (ms-1):
Final Velocity v (ms-1):
Acceleration a (ms-2):
Time t (s):
Displacement s (m):
Current Displacement s: 0m
Current Time t: 0s
Animation Speed:
Set positive as:
Right
Upwards
Solution 1
Solution 2
Solution:
SOLUTION NOT VALID AS TIME t < 0.
Steps for calculating this are still shown below for reference
but remember this is not a valid solution so please check your
entered values again.
Four Equations of Motion
v = u + at
v2 = u2 + 2as
s = ut + 1/2 at2
s = 1/2 (u+v) t
Explanation and Steps
Generally, you know the values of 3 variables. First look for the equation that
contains all 3 variables that you know in it. Each equation of motion has exactly 4 variables,
if you know 3 of the variables you can calculate the 4th variable inside this equation directly.
Choose the equation that is most easy to manipulate, for example you should avoid equations with
squares if possible unless you have no other choice.
Notice that v can take on 2 values, one positive and one negative as it is the square root of u2 + 2as.
(Recall that squaring a number and its negative will give the same result.)
The other value of v is in the other solution which you can view by clicking on the corresponding checkbox above.
Each of the solutions will give rise to a t value each. One of the solutions might not be valid if the
t calculated in the next step is negative as time cannot be negative. If both values of v give positive t then
there are 2 valid solutions.
Notice that u can take on 2 values, one positive and one negative as it is the square root of v2 - 2as.
(Recall that squaring a number and its negative will give the same result.)
The other value of u is in the other solution which you can view by clicking on the corresponding checkbox above.
Each of the solutions will give rise to a t value each. One of the solutions might not be valid if the
t calculated in the next step is negative as time cannot be negative. If both values of u give positive t then
there are 2 valid solutions.
So now you have 4 variables. Now look for an equation that contains
the final variable that you still do not know. The values of the other 3 variables in this equation
should be something you know already as you only have one unknown variable at this point. Again try
to look for the equation that is easiest to manipulate.
Physically what does it mean to have 2 valid solutions of v and t for the same u, a and s?
The object starts off with initial velocity u and continues travelling with acceleration a. On the way it passes by
the point with displacement s with a certain velocity v1,
at a certain time t1. It travels further before turning back to the same point at the same displacement s,
at a later time t2 with a velocity v2. But this time it is travelling in the opposite direction
as the previous time when it passes through this point as it has turned backward and changed direction. Hence this v2
is not the same as the previous v1.
Acceleration a is the same throughout the whole motion and the object passes through
the same point of displacement s at two different times and two different velocities. Hence for the same u, a and s
there are 2 sets of solutions: v1,t1 and v2, t2. Play the animation for both solutions to
see how the same point is reached at 2 different times.
Physically what does it mean to have 2 valid solutions of u and t for the same v, a and s?
In one of the solutions, the object starts off with initial velocity u1 and travels towards the point at displacement s at an acceleration a in the same direction.
It reaches displacement s at t1 with final velocity v. In the other solution, the object starts off with a different initial velocity
u2 which moves away from the point at displacement s at first. (Meaning u2 is in the opposite direction and sign as u1.)
It travels with the same acceleration a directed towards s, hence at some point it would slow down and change direction to move towards displacement s.
It will reach displacement s after passing through its starting point, at time t2 but with same final velocity v1
In both cases the object reaches the same displacement s with 2 different start initial velocities u, and arrive at two different times.
Hence there are 2 sets of solutions: u1,t1 and u2, t2. Play the animation for both solutions to
see how the same point is reached at 2 different times but with 2 different initial velocities u in opposite directions.
Damian's Physics Simulations
Email: damianboh.ck@gmail.com
