
The student will solve example questions related to uniformly accelerated motion. (B2.7, B3.3)
“It goes from zero to 60 in about
3 seconds.”

SPH4C

Acceleration
Acceleration is defined as the change in the velocity of an object per interval of time, or:

Acceleration
Acceleration is defined as the change in the velocity of an object per interval of time, or: a


 v

 t

Acceleration
Acceleration is defined as the change in the velocity of an object per interval of time, or:
 a

 v

 t
 v

2

 t
 v
1

Acceleration
Acceleration is defined as the change in the velocity of an object per interval of time, or:
 a

 v

 t
 v

2

 t
 v
1 v
2
= final velocity and v
1
= initial velocity

Acceleration
 a

 v

 t
 v

2

 t
 v
1
Acceleration will have units of m/s/s or m/s 2 .
(Notice also that acceleration is a vector.)

Example 1
Jake increases his velocity from 13 m/s [E] to 25 m/s [E] in 5.0 s. What is his acceleration?

Example 1
Jake increases his velocity from 13 m/s [E] to 25 m/s [E] in 5.0 s. What is his acceleration?
Givens v
 v

1
2


13
25 m
: s m s
[
[
E
E
]
]
 t

5 .
0 s
Unknown
 a

?
:

Example 1
Jake increases his velocity from 13 m/s [E] to 25 m/s [E] in 5.0 s. What is his acceleration?
Givens : v
1
 
13 m s v
2
 t



25
5 .
0 s m s
Unknown : a

?

Example 1
Jake increases his velocity from 13 m/s [E] to 25 m/s [E] in 5.0 s. What is his acceleration?
Givens : v
1
 
13 m s v
2
 t



25
5 .
0 s m s
Unknown : a

?
a
 a
 v
2

 t v
1
25 m s

13 m s
5 .
0 s

2 .
4 m s
2

Example 1
Jake increases his velocity from 13 m/s [E] to 25 m/s [E] in 5.0 s. What is his acceleration?
Givens : v
1
 
13 m s v
2
 t



25
5 .
0 s m s
Unknown : a

?
a
 a
 v
2

 t v
1
25 m s

13 m s
5 .
0 s

2 .
4 m s
2
His acceleration is 2.4 m/s 2 [E].

Example 2
Richard reduces his velocity from 15 m/s [forward] to
12 m/s [forward] in 4.0 s. What is his acceleration?

Example 2
Richard reduces his velocity from 15 m/s [forward] to
12 m/s [forward] in 4.0 s. What is his acceleration?
Givens : v
1
 
15 m s v
2
 t



12
4 .
0 s m s
Unknown : a

?

Example 2
Richard reduces his velocity from 15 m/s [forward] to
12 m/s [forward] in 4.0 s. What is his acceleration?
Givens v
1
 
15
: m s v
2
 t



12
4 .
0 s m s
Unknown : a

?
a a

 v
2

 t v
1
12 m s

15 m s
4 .
0 s
 
0 .
75 m s
2

Example 2
Richard reduces his velocity from 15 m/s [forward] to
12 m/s [forward] in 4.0 s. What is his acceleration?
Givens v
1
 
15
: m s v
2
 t



12
4 .
0 s m s
Unknown : a

?
a a

 v
2

 t v
1
12 m s

15 m s
4 .
0 s
 
0 .
75 m s
2
His acceleration is 0.75 m/s 2
[backward].

Rearranging the Equation
The equation that represents the definition of acceleration may be rearranged to solve for any of the other variables, e.g.
(solving for v
2
) a
 a
 t v
2

 t v
1
 v
2

 t v
1
 t a
 t
 v
2
 v
1 a
 t
 v
1
 v
2
 a
 t or v
2

 v
1 v
1

 v
2 a
 t v
1
 v
1

Rearranging the Equation
The equation that represents the definition of acceleration may be rearranged to solve for any of the other variables, e.g.
a
 a
 t v
2

 t v
1
 v
2

 t v
1
 t a
 t
 v
2
 v
1 a
 t
 v
1
 v
2
 a
 t or v
2

 v
1 v
1

 v
2 a
 t v
1
 v
1

Rearranging the Equation
The equation that represents the definition of acceleration may be rearranged to solve for any of the other variables, e.g.
a
 a
 t v
2

 t v
1
 v
2

 t v
1
 t a
 t
 v
2
 v
1 a
 t
 v
1
 v
2
 a
 t or v
2

 v
1 v
1

 v
2 a
 t v
1
 v
1

Rearranging the Equation
The equation that represents the definition of acceleration may be rearranged to solve for any of the other variables, e.g.
a
 a
 t v
2

 t v
1
 v
2

 t v
1
 t a
 t
 v
2
 v
1 a
 t
 v
1
 v
2
 a
 t or v
2

 v
1 v
1

 v
2 a
 t v
1
 v
1

Rearranging the Equation
The equation that represents the definition of acceleration may be rearranged to solve for any of the other variables, e.g.
a
 a
 t v
2

 t v
1
 v
2

 t v
1
 t a
 t
 v
2
 v
1 a
 t
 v
1
 v
2
 a
 t or v
2

 v
1 v
1

 v
2 a
 t v
1
 v
1

Rearranging the Equation
The equation that represents the definition of acceleration may be rearranged to solve for any of the other variables, e.g.
a
 a
 t v
2

 t v
1
 v
2

 t v
1
 t a
 t
 v
2
 v
1 a
 t
 v
1
 v
2
 a
 t or v
2

 v
1 v
1

 v
2 a
 t v
1
 v
1

Example 3
A ball is travelling at 12 m/s [up] and being accelerated at 9.8 m/s 2 [down]. What is its velocity after 2.0 s?
Givens : v
1 a

 
12 m s

9 .
8 m s
2
 t

2 .
0 s
Unknown : v
2

?

Example 3
A ball is travelling at 12 m/s [up] and being accelerated at 9.8 m/s 2 [down]. What is its velocity after 2.0 s?
Givens : v
1 a

 
12 m s

9 .
8 m s
2
 t

2 .
0 s
Unknown : v
2

?
Select : a
Solve : v
2

 v
2

 t v
1
12 m s


(

9 .
8 v
2
 v
1 m s
2
)( 2 .
0 s )
 a
 t
 
7 .
6 m s

Example 3
A ball is travelling at 12 m/s [up] and being accelerated at 9.8 m/s 2 [down]. What is its velocity after 2.0 s?
Givens : v
1 a

 
12 m s

9 .
8 m s
2
 t

2 .
0 s
Unknown : v
2

?
Select : a
Solve : v
2

 v
2

 t v
1
12 m s


(

9 .
8 v
2
 v
1 m s
2
)( 2 .
0 s )
 a
 t
 
7 .
6 m s

Example 3
A ball is travelling at 12 m/s [up] and being accelerated at 9.8 m/s 2 [down]. What is its velocity after 2.0 s?
Givens : v
1 a

 
12 m s

9 .
8 m s
2
 t

2 .
0 s
Unknown : v
2

?
Select : a
Solve : v
2

 v
2

 t v
1
12 m s


(

9 .
8 v
2
 v
1 m s
2
)( 2 .
0 s )
 a
 t
 
7 .
6 m s
Its velocity is 7.6 m/s [down].

More Practice
Playing with the Ticker Timers and Acceleration
Graphing Acceleration