Motion in a straight line under constant acceleration is a common physics homework problem. The equations of motion to describe these conditions that can be used to solve any problem associated with them. These equations are:
(1) x = x0 + v0t + ½at2
(2) v = v0 + at
(3) v2 = v02 + 2a(x – x0)
x is the distance travelled
x0 is the initial starting point
v is the velocity
v0 is the initial velocity
a is the acceleration
t is the time
This example problem shows how to use these equations to calculate position, velocity and time of a constantly accelerating body.
A block slides along a frictionless surface with a constant acceleration of 2 m/s2. At time t = 0 s the block is at x = 5m and travelling with a velocity of 3 m/s.
a) Where is the block at t = 2 seconds?
b) What is the block’s velocity at 2 seconds?
c) Where is the block when it’s velocity is 10 m/s?
d) How long did it take to get to this point?
Here is an illustration of the setup.
The variables we know are:
x0 = 5 m
v0 = 3 m/s
a = 2 m/s2
Part a) Where is the block at t = 2 seconds?
Equation 1 is the useful equation for this part.
x = x0 + v0t + ½at2
Substitute t = 2 seconds for t and the appropriate values of x0 and v0.
x = 5 m + (3 m/s)(2 s) + ½(2 m/s2)(2 s)2
x = 5 m + 6 m + 4 m
x = 15 m
The block is at the 15 meter mark at t = 2 seconds.
Part b) What is the block’s velocity at t = 2 seconds?
This time, Equation 2 is the useful equation.
v = v0 + at
v = (3 m/s) + (2 m/s2)(2 s)
v = 3 m/s + 4 m/s
v = 7 m/s
The block is travelling 7 m/s at t = 2 seconds.
Part c) Where is the block when it’s velocity is 10 m/s?
Equation 3 is the most useful at this time.
v2 = v02 + 2a(x – x0)
(10 m/s)2 = (3 m/s)2 + 2(2 m/s2)(x – 5 m)
100 m2/s2 = 9 m2/s2 + 4 m/s2(x – 5 m)
91 m2/s2 = 4 m/s2(x – 5 m)
22.75 m = x – 5 m
27.75 m = x
The block is at the 27.75 m mark.
Part d) How long did it take to get to this point?
There are two ways you could do this. You could use Equation 1 and solve for t using the value you calculated in part c of the problem, or you could use equation 2 and solve for t. Equation 2 is easier.
v = v0 + at
10 m/s = 3 m/s + (2 m/s2)t
7 m/s = (2 m/s2)t
7⁄2 s = t
It takes 7⁄2 s or 3.5 s to get to the 27.75 m mark.
One tricky part of this type of problem is you have to pay attention to what the question is asking for. In this case, you were not asked how far the block travelled, but where it is. The reference point is 5 meters from the origin point. If you needed to know how far the block travelled, you would have to subtract the 5 meters.
For further help, try these Equations of Motion example problems:
Equations of Motion – Interception Example
Equations of Motion – Vertical Motion
Equations of Motion – Breaking Vehicle
Equations of Motion – Projectile Motion
Last modified: August 19th, 2014 by