Page 50 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 3] MOTION IN A STRAIGHT LINE 35

v 0 + v

and so s = ¯vt = t
2
Since v = v 0 + at, another way to specify the distance covered during t is

v 0 + v 0 + at 1
s = t = v 0 t + at 2
2
2
If the body is accelerated starting from rest, v 0 = 0 and
1
s = at 2
2
Another useful formula gives the final velocity of a body in terms of its initial velocity, its acceleration, and
the distance it has traveled during the acceleration:

2
2
v = v + 2as
0
This can be solved for the distance s to give
2
v − v 0 2
s =
2a
In the case of a body that starts from rest, v 0 = 0 and
√ v 2
v = 2as s =
2a
Table 3.1 summarizes the formulas for motion under constant acceleration.

Table 3-1 Formulas for Motion
under Constant Acceleration
Distance Final Velocity

v 0 + v

s = t v = v 0 + at
2
1
2
2
s = v 0 t + at 2 v = v + 2as
2 0
SOLVED PROBLEM 3.10
2
2
Derive the formula v = v + 2as.
0
1
2
We start with the formula s = v 0 t + at and substitute for t the expression
2
v − v 0
t =
a
The result is
1
s = v 0 t + at 2
2
(v − v 0 ) 1 (v − v 0 ) 2
= v 0 + a 2
a 2 a
v 0 v v 2 v 2 v 0 v v 2
= − 0 + − + 0
a a 2a a 2a
2
v − v 2
= 0
2a
Multiplying through by 2a and rearranging gives
2
2
v = v + 2as
0