The simplest type of accelerated motion is motion in a straight line and constant acceleration. The velocity changes at the same rate as the motion progresses. There are three basic equations of motion that will get you through most homework problems that deal with motion in a straight line at constant acceleration.
(1) x = x0 + v0t + ½at2
(2) v = v0 + at
(3) v2 = v02 + 2a(x – x0)
where
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 worked constant acceleration example problem will show how to use these three equations of motion to find details about the position, velocity and acceleration of a breaking vehicle.
Example Problem:
A motorist is speeding along at 120 km/hr when he sees a squirrel on the road 200 meters in front of him. He tries to stop, but it takes 12 seconds for his car to stop.
(a) What is the acceleration of the car? (assume acceleration was constant)
(b) Does the squirrel survive?
(c) How fast was the car moving at 100 meters?
Solution:
This shows the conditions of the vehicle at the beginning (t = 0 s) and when the car has come to a stop (v = 0 km/hr)
Part a) Find the acceleration.
Use Equation 2 from above.
v = v0 + at
Use t = 12 seconds and v0 = 120 km/hr. First, notice the stopping time is in seconds, but the velocity is per hour. We will also need the distance to be in meters, so convert the velocity to m/s:
v0 = 33.33 m/s
The car is stopped at the end, so the final velocity is equal to zero.
0 = 33.33 m/s + a(12 s)
-33.33 m/s = a(12 s)
a = -2.78 m/s2
Note the acceleration is negative. This means it is slowing down the vehicle as motion progresses in the positive direction. Just what you would expect in a problem where the vehicle is slowing down.
Part b) Does the squirrel survive?
In order to find out if the squirrel survives, we need to know how far the vehicle travelled before it stops. If the distance travelled is less than the distance to the squirrel, the squirrel will survive. Use equation 1 from above for this part.
x = x0 + v0t + ½at2
Plug in the acceleration from part a) and the initial conditions.
x = 0 m + (33.33 m/s)(12 s) + ½(-2.78 m/s2)(12 s)2
x = 399.96 m – 200.16 m
x = 199.8 m
The distance the vehicle took to stop was less than 200 m, so the squirrel did survive the encounter…barely.
Part c) What was the velocity at 100 meters?
Use equation 3 from above for this part.
v2 = v02 + 2a(x – x0)
Use x = 100 meters and the units of the velocities in m/s and acceleration in m/s2.
v2 = (33.33 m/s)2 + 2(-2.78 m/s2)(100 m – 0 m)
v2 = 1110.89 m2/s2 – 556 m2/s2
v2 = 554.89 m2/s2
v = 23.56 m/s
The vehicle was moving at 23.56 m/s (84.8 km/hr) at the 100 meter mark.