Acceleration

review Honors

Book #33

A stone is dropped from the top of a cliff.

It hits the ground below after 3.25s. How

high is the cliff?

Book #33 A stone is dropped from the top of a cliff. It hits the

ground below after 3.25s. How high is the cliff?

Given:

a = 9.8 m/s2

t = 3.25s

vi = 0 m/s

Δx = ?

Δx = vi*t + 0.5*a*t2

Δx = (0 m/s)*(3.25 s)+ 0.5*(9.8 m/s2)*(3.25 s)2

Δx = 51.76m

Book #34

If a car rolls gently (vo = 0) off of a vertical

cliff, how long does it take it to reach 85

km/hr?

Book #34 If a car rolls gently (vo = 0) off of a vertical cliff, how

long does it take it to reach 85 km/hr?

Given:

a = 9.8 m/s2

t = ?

vi = 0 m/s

Vf = 85 km/hr

Step 1 convert

85km/hr = m/s

23.61 m/s

Step 2

Vf = vi + at

23.61 = 0 + 9.8(t)

2.41s = t

Book #37

A ball player catches a ball 3.0s after

throwing it vertically upward. With what

speed did he throw it, and what height

did it reach?

Book #37 A ball player catches a ball 3.0s after throwing it

vertically upward. With what speed did he throw it,

and what height did it reach?

Given:

a = -9.8 m/s2

t = 3.0 s

vi = ?

Δx = ?

Step 1

Vf = vi + at

0 = vi + at

14.7m/s = vi

Step 2

Δx = vi*t + 0.5*a*t2

Δx = (14.7 m/s)*(1.5 s)+ 0.5*(-9.8 m/s2)*(1.5 s)2

Δx = 11.03m

Problem 1

An airplane accelerates down a runway

at 3.20 m/s2 for 32.8 s until is finally lifts off

the ground. Determine the distance

traveled before takeoff.

Problem 1 An airplane accelerates down a runway at 3.20

m/s2 for 32.8 s until is finally lifts off the ground.

Determine the distance traveled before takeoff.

Given:

a = +3.2 m/s2

t = 32.8 s

vi = 0 m/s

Δx = ?

Δx = vi*t + 0.5*a*t2

Δx = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2

Δx = 1721.43 m

Problem 2

A car starts from rest and accelerates

uniformly over a time of 5.21 seconds for a

distance of 110 m. Determine the

acceleration of the car.

Problem 2 A car starts from rest and accelerates uniformly over

a time of 5.21 seconds for a distance of 110 m.

Determine the acceleration of the car.

Given:

Δx = 110 m

t = 5.21 s

vi = 0 m/s

a = ?

Δx = vi*t + 0.5*a*t2

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2

110 m = (13.57 s2)*a

a = (110 m)/(13.57 s2)

a = 8.10 m/s2

Problem 3

A race car accelerates uniformly from

18.5 m/s to 46.1 m/s in 2.47 seconds.

Determine the acceleration of the car

and the distance traveled.

Problem 3 A race car accelerates uniformly from 18.5 m/s to

46.1 m/s in 2.47 seconds. Determine the

acceleration of the car and the distance traveled.

Given:

vi = 18.5 m/s

vf = 46.1 m/s

t = 2.47 s

Δx = ??

a = ??

a = (Delta v)/t

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.17 m/s2

d = vi*t + 0.5*a*t2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2

d = 45.7 m + 34.1 m

d = 79.77 m

(Note: can use vf2 = vi2 + 2*a*d)

Problem 4

A baseball is popped straight up into the

air and has a hang-time of 6.25 s.

Determine the height to which the ball

rises before it reaches its peak.

Problem 4 A baseball is popped straight up and is in the air for

6.25 s. Determine the height to which the ball rises

before it reaches its peak.

Given:

a = -9.8 m/s2

vf = 0 m/s

t = 3.13 s

Δx = ??

Step 1: vf = vi + a*t

0 m/s = vi + (-9.8 m/s2)*(3.13 s)

0 m/s = vi - 30.6 m/s

vi = 30.63 m/s

Step 2: vf2 = vi2 + 2*a*d

(0 m/s)2 = (30.6 m/s)2 + 2*(-9.8 m/s2)*(d)

0 m/s2 = (937.89 m/s) + (-19.6 m/s2)*d

-937.89 m/s = (-19.6 m/s2)*d

(-937.89 m/s)/(-19.6 m/s2) = d

d = 47.85 m

Problem 5

A stone is dropped into a deep well and is

heard to hit the water 3.41 s after being

dropped. Determine the depth of the

well.

Problem 5 A stone is dropped into a deep well and is heard to

hit the water 3.41 s after being dropped. Determine

the depth of the well.

Given:

a = -9.8 m/s2

t = 3.41 s

vi = 0 m/s

Δx = ??

Δx = vi*t + 0.5*a*t2

Δx = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s2)*(3.41 s)2

Δx = 0 m+ 0.5*(-9.8 m/s2)*(11.63 s2)

Δx = -56.99 m

(NOTE: the - sign indicates direction)

Problem 6

How long does it take a ball to reach the

ground 7.0 m below, if it is thrown straight

up with an initial speed of 2.00 m/s?

Problem 6 How long does it take a ball to reach the ground

7.0 m below, if it is thrown straight up with an initial

speed of 2.00 m/s?

Given:

Δx = 7.0 m

a= -9.8 m/s2

vi = 2.0 m/s *

t =?

Δx = vit + (0.5)at2

(7.0 m) = (-2.00)t + (0.5)(9.8)t2

t = {1.42, -1.01} Since t < 0 has no meaning,

t = 1.42 s

Use Quadratic Formula to solve,

x = -b ±√( b2 – 4ac)

2a

Book #47 A stone is thrown vertically upward with a speed

of 12.0 m/s from the edge of a cliff 70.0m high. How much later does it reach the bottom of the cliff?

Given:

a = -9.8m/s2

vi = 12.0 m/s

Δx = -70 m

t = ?

Δx = vi*t + ½ *a*t2

-70 = (12)*(t)+ 0.5*(-9.8)*(t)2

-70 = 12t + -4.9t2

t = {5.20, -2.74}

Since t < 0 has no meaning,

t = 5.20 s

Use Quadratic Formula to solve,

x = -b ±√( b2 – 4ac)

2a

Book #47

A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 70.0m high. a. What is its speed just before hitting?

Given:

a = -9.8m/s2

vi = 12.0 m/s

Δx = -70 m

t = 5.20

Vf = ?

Vf = vi + at

Vf = 12 + (-9.8)(5.20)

Vf = -38.9 m/s

Book #47 A stone is thrown vertically upward with a speed of

12.0 m/s from the edge of a cliff 70.0m high.

b. Determine the distance the stone traveled

Given:

a = -9.8m/s2

vi = 12.0 m/s

Δx = -70 m (only includes height of cliff)

t = 5.20s

Vf = -38.9m/s

Vf2 = vi2 + 2ad

02 = (12)2 + 2(-9.8)d

-144 = -19.6d

D = 7.35m

so 7.35m + 70m = 77.35m total distance