The third equation is derived from the first and the second equations. The mathematics is straight forward as I will show below.
To derive the third law we are going to use the first equation of motion as shown below
v = u +at Equation 1
and this second equation that we used when we derived the second equation. If you have forget check the post on the second equation of motion.
s = 1/2( v + u)*t Equation 2
Now if we change the first equation of motion we will have
We can then substitute equation 3 above in equation 2 to obtain
Multiplying on both side by 2a
So this is the derivation for the third equation of motion.
To answer this question we first have to identify the different variables.
distance travelled s = ???
initial speed u = 10 m/s
final speed v = o m/s
acceleration = - 2 m/s2
Note that the acceleration is negative since deceleration involves a decrease in speed hence the number is positive.
We now need to use the third equation of motion as shown below.
Substituting the different values in the equation
2*-2*s = o2 - 102
-4s = –100
s =25 m
Hence the distance travelled by the object during the deceleration is 25 m