Acceleration

A car starts from rest and accelerates at a
rate of 6.1 m/s2 for 7 seconds. Calculate
how far it traveled.
 V1 =
 V2 =
A=
D=
T=
Acceleration

A car starts from rest and accelerates at a
rate of 6.1 m/s2 for 7 seconds. Calculate
how far it traveled.
 V1 = 0 m/s
Acceleration

A car starts from rest and accelerates at a
rate of 6.1 m/s2 for 7 seconds. Calculate
how far it traveled.
 V1 = 0 m/s
 V2 = no
Acceleration

A car starts from rest and accelerates at a
rate of 6.1 m/s2 for 7 seconds. Calculate
how far it traveled.
 V1 = 0 m/s
 V2 = no
 A = 6.1 m/s2
Acceleration

A car starts from rest and accelerates at a
rate of 6.1 m/s2 for 7 seconds. Calculate
how far it traveled.
 V1 = 0 m/s
 V2 = no
 A = 6.1 m/s2
D=?
Acceleration

A car starts from rest and accelerates at a
rate of 6.1 m/s2 for 7 seconds. Calculate
how far it traveled.
 V1 = 0 m/s
 V2 = no
 A = 6.1 m/s2
D=?
 T = 7 sec.
Solution #1

D = V1t + at2
2
Solution #1
D = ½ at2
 D = (6.1 m/s2)(7 sec)2
2

Solution #1
D = ½ at2
 D = (6.1 m/s2)(7 sec)2
2
 D = 150 m

Acceleration

#2
What was the initial velocity of a car that
accelerates at a rate of – 8.0 m/s2 and comes
to rest in 484 m?
 V1=
 V2=
A=
D=
T=
Acceleration

#2
What was the initial velocity of a car that
accelerates at a rate of – 8.0 m/s2 and comes
to rest in 484 m?
 V1 = ?
Acceleration

#2
What was the initial velocity of a car that
accelerates at a rate of – 8.0 m/s2 and comes
to rest in 484 m?
 V1 = ?
 V2 = 0 m/s
Acceleration

#2
What was the initial velocity of a car that
accelerates at a rate of – 8.0 m/s2 and comes
to rest in 484 m?
 V1 = ?
 V2 = 0 m/s
 A = -8.0 m/s2
Acceleration

#2
What was the initial velocity of a car that
accelerates at a rate of – 8.0 m/s2 and comes
to rest in 484 m?
 V1 = ?
 V2 = 0 m/s
 A = -8.0 m/s2
 D = 484 m
Acceleration

#2
What was the initial velocity of a car that
accelerates at a rate of – 8.0 m/s2 and comes
to rest in 484 m?
 V1 = ?
 V2 = 0 m/s
 A = -8.0 m/s2
 D = 484 m
 T = no
Solution #2

Use equation # 7
 V22 = V12 + 2ad
Solution #2

Use equation # 7
 V22 = V12 + 2ad
 0 m/s = V12 + 2(-8.0m/s)( 484 m)
Solution #2

Use equation # 7
 V22 = V12 + 2ad
 0 m/s = V12 + 2(-8.0m/s2)( 484 m)
 7744 m2/s2 = V12
Solution #2

Use equation # 7
 V22 = V12 + 2ad
 0 m/s = V12 + 2(-8.0m/s2)( 484 m)
 7744 m2/s2 = V12
 V1= 88 m/s
Acceleration #3

What is the acceleration of a car that was
going at 25.6 m/s and slows to 8.5 m/s in
4.6 sec.?
 V1=
 V2=
 A=
 D=
 T=
Acceleration #3

What is the acceleration of a car that was
going at 25.6 m/s and slows to 8.5 m/s in
4.6 sec.?
 V1 = 25.6 m/s
Acceleration #3

What is the acceleration of a car that was
going at 25.6 m/s and slows to 8.5 m/s in
4.6 sec.?
 V1 = 25.6 m/s
 V2 = 8.5 m/s
Acceleration #3

What is the acceleration of a car that was
going at 25.6 m/s and slows to 8.5 m/s in
4.6 sec.?
 V1 = 25.6 m/s
 V2 = 8.5 m/s
A=?
Acceleration #3

What is the acceleration of a car that was
going at 25.6 m/s and slows to 8.5 m/s in
4.6 sec.
 V1 = 25.6 m/s
 V2 = 8.5 m/s
A=?
 D = no
Acceleration #3

What is the acceleration of a car that was
going at 25.6 m/s and slows to 8.5 m/s in
4.6 sec.
 V1 = 25.6 m/s
 V2 = 8.5 m/s
A=?
 D = no
 T = 4.6 sec
Solution #3

Use Equation #2
 A = V2- V1
t
Solution #3

A = V2- V1
t
A = 8.5 m/s – 25.6 m/s
4.6 sec
Solution #3

A = V2- V1
t
A = 8.5 m/s – 25.6 m/s
4.6 sec
A = -3.7 m/s2
Acceleration #4

A car traveling at 14m/s encounters a patch
of ice and takes 5.0 sec to stop. How far did
it skid?
 V1 =
 V2=
A=
D=
T=
Acceleration #4

A car traveling at 14m/s encounters a patch
of ice and takes 5.0 sec to stop. How far did
it skid ?
 V 1 = 14 m/s
Acceleration #4

A car traveling at 14m/s encounters a patch
of ice and takes 5.0 sec to stop. How far did
it skid ?
 V 1 = 14 m/s
 V2 = 0 m/s
Acceleration #4

A car traveling at 14m/s encounters a patch
of ice and takes 5.0 sec to stop. How far did
it skid ?
 V 1 = 14 m/s
 V2 = 0 m/s
 A = no
Acceleration #4

A car traveling at 14m/s encounters a patch
of ice and takes 5.0 sec to stop. How far did
it skid ?
 V 1 = 14 m/s
 V2 = 0 m/s
 A = no
D=?
Acceleration #4

A car traveling at 14m/s encounters a patch
of ice and takes 5.0 sec to stop. How far did
it skid ?
 V 1 = 14 m/s
 V2 = 0 m/s
 A = no
D=?
 T = 5.0 sec
Solution #4

Use equation # 4
 D = (V2+V1) t
2
Solution #4

Use equation # 4
 D = (V2+V1) t
2
D = (14m/s + 0 m/s)5 sec
2
Solution #4

Use equation # 4
 D = (V2+V1) t
2
D = (14m/s + 0 m/s)5 sec
2
D = 35 m
Acceleration #5

An object traveling at 16 m/s accelerates at
a constant rate of 4.0 m/s2 over 50 meters.
What is it’s final velocity?
 V1=
 V2=
 a=
 d=
 t=
Acceleration #5

An object traveling at 16 m/s accelerates at
a constant rate of 4.0 m/s2 over 50 meters.
What is it’s final velocity?
 V1= 16 m/s
Acceleration #5

An object traveling at 16 m/s accelerates at
a constant rate of 4.0 m/s2 over 50 meters.
What is it’s final velocity?
 V1= 16 m/s
 V2 = ?
Acceleration #5

An object traveling at 16 m/s accelerates at
a constant rate of 4.0 m/s2 over 50 meters.
What is it’s final velocity?
 V1= 16 m/s
 V2 = ?
 A = 4.0 m/s2
Acceleration #5

An object traveling at 16 m/s accelerates at
a constant rate of 4.0 m/s2 over 50 meters.
What is it’s final velocity?
 V1= 16 m/s
 V2 = ?
 A = 4.0 m/s2
 D = 50 m
Acceleration #5

An object traveling at 16 m/s accelerates at
a constant rate of 4.0 m/s2 over 50 meters.
What is it’s final velocity?
 V1= 16 m/s
 V2 = ?
 A = 4.0 m/s2
 D = 50 m
 T = no
Solution #5

Use equation #7
 V22 = V12 + 2ad
Solution #5

Use equation #7
 V22 = V12 + 2ad
 V22 = 16 m/s + 2(4.0m/s2)(50m)
Solution #5

Use equation #7
 V22 = V12 + 2ad
 V22 = 16 m/s + 2(40.0m/s2)(50m)
 V2 = 26 m/s