Problem 16

A car traveling initially at +7.0 m/s accelerates at a rate of +0.80 m/s2 for an interval of 2.0 s.What is the velocity at the end of the acceleration?

SOLUTION:

Step 1 is to identify the knowns and unknowns.
 

KNOWNS

UNKNOWN

v0 = +7.0 m/s

v

a = +0.80 m/s2

 

t = 2.0 s

 

Step 2 is to identify which equation is best to use based on the known and unknown information in the problem.The best equation is:

v = v0 + at = (7.0 m/s) + (0.80 m/s2)(2.0s)

v = 8.6 m/s

Problem 20

The velocity versus time graph for an object moving along a straight path shown in the figure below.

(a) Find the average acceleration of this object during the time intervals 0 to 5.0 s, 5.0 s to 15 s, and 0 to 20 s.

SOLUTION:

Average acceleration a = Dv/Dt = (vf – vi)/(tf – ti)

First Interval: 0 to 5.0 s

                        vf = -8.0 m/s

                        vi = -8.0 m/s

                        a = 0

Second Interval: 5.0 s to 15 s

                        vf = 8.0 m/s

                        vi = -8.0 m/s

                        Dv = 8.0 – (-8.0) = 16.0 m/s

                        a = (16.0 m/s)/(10 s)

                        a = 1.6 m/s2

Third Interval: 0 s to 20 s

                        vf = 8.0 m/s

                        vi = -8.0 m/s

                        Dv = 8.0 – (-8.0) = 16.0 m/s

                        a = (16.0 m/s)/(20 s)

                        a = 0.80 m/s2

(b) Find the instantaneous acceleration at 2.0, 10, and 18 s.

SOLUTION:

The slope is zero from 0 to 5.0 s therefore the instantaneous acceleration at 2.0 s is ZERO

The slope is 1.6 m/s2 between 5.0 s and 15 s therefore the instantaneous acceleration a = 1.6 m/s2 at t = 10 s.

The slope is zero from 15 s to 20 s therefore the instantaneous acceleration at 18 s is ZERO

 

Problem 24

A jet plane lands with a velocity of +100 m/s and can accelerate at a maximum rate of –5.0 m/s2 as it comes to rest.

(a) At the instant it touches the runway, what is the minimum time needed before it can come to rest?

SOLUTION:

Step 1 is to identify the knowns and unknowns.
 

KNOWNS

UNKNOWN

v0 = +100 m/s

t

a = -5.0 m/s2

 

v = 0

 

Step 2 is to identify which equation is best to use based on the known and unknown information in the problem.The best equation is:

v = v0 + at

t = (v – v0)/a = (0 – 100 m/s)/(-5.0 m/s2) = 20 s

(b) Can this plane land on a small island airport where the runway is 0.80 km long?

SOLUTION:

To do this, we need to find the distance traveled in the time calculated above.
 

KNOWNS

UNKNOWN

v0 = +100 m/s

x

a = -5.0 m/s2

 

v = 0

 

x0 = 0

 

t = 20 s

 

x = x0 + ½ (v + v0)t

x = 0 + ½ (0 + 100 m/s)(20 s)

x = 1000 m = 1.0 km

Therefore is cannot land on this small island airport.

 

Problem 29

A speedboat increases in speed uniformly from 20 m/s to 30 m/s in a distance of 200 m.

(a) Find the magnitude of its acceleration.

SOLUTION:
 

KNOWNS

UNKNOWN

v0 = 20 m/s

a

v = 30 m/s

 

x0 = 0

 

x = 200 m

 

(b) Find the time it takes the boat to travel this distance.
 

KNOWNS

UNKNOWN

v0 = 20 m/s

t

v = 30 m/s

 

x0 = 0

 

x = 200 m

 

a = 1.25 m/s2

 

      v = v0 + at

      t = (v – v0)/a= (30 – 20)/(1.25)

t = 8.0 s

 

Problem 34

A car starts from rest and travels 5.0 s with a uniform acceleration of 1.5 m/s2.The driver then applies the brakes, causing a uniform acceleration of –2.0 m/s2.If the brakes are applied for 3.0 s, how fast is the car going at the end of the brake period and how far has it gone?

SOLUTION:

Break this problem up into parts.

The first part is the motion where the car starts from rest and travels 5.0 s while accelerating 1.5 m/s2.The first thing we must find is the velocity of the car at the end on the 5.0 s.
 

KNOWNS

UNKNOWN

v0 = 0

v

a = 1.5 m/s2

 

x0 = 0

 

t = 5.0 s

 

v = v0 + at = 0 + (1.5 m/s2)(5.0 s) = 7.5 m/s

The second part involves looking for the distance traveled in 5.0 s.
 

KNOWNS

UNKNOWN

v0 = 0

x

v = 7.5 m/s

 

t = 5.0 s

 

x0 = 0

 

x = x0 + ½ (v + v0)t = 0 + ½ (0 + 7.5 m/s)(5.0 s)

x = 18.75 m

The third part of the problem is when the driver applies the brakes and slows down.
 

KNOWNS

UNKNOWN

v0 = 7.5 m/s

v

a = -2.0 m/s2

 

t = 3.0 s

 

v = v0 + at = (7.5 m/s) + (-2.0 m/s2)(3.0 s) = 1.5 m/s

The fourth part involves finding the distance from the original point of origin (x = 0).
 

KNOWNS

UNKNOWN

v0 = 7.5 m/s

x

a = -2.0 m/s2

 

t = 3.0 s

 

x0 = 18.75 m

 

v = 1.5 m/s

 

x = x0 + ½ (v + v0)t = 18.75 m + ½ (1.5 m/s + 7.5 m/s)(3.0 s)

x = 18.75 m + 13.5 m

x = 32.3 m

 

Problem 36

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 12 m/s skates by with the puck.After 3.0 s the first player decides to chase the second player.If he accelerates uniformly at 4.0 m/s2,

(a) How long does it take him to catch his opponent?

SOLUTION:

At t = 0, both players are at the origin.At t = 3.0 s, player 1 is at the origin and player 2 has skated away.Find the distance between the two players after 3 seconds and write an equation for each player.

Step 1: Player 2 is not accelerating and his speed is 12 m/s, therefore

x = vt = (12 m/s)(3.0 s) = 36 m

Step 2: Write an equation for each player.

                        Player 1: x(1) = x0 + v0t + ½ at2 = 0 + 0 + ½ (4.0 m/s2)t = (2.0 m/s2)t2

                        Player 2: x(2) = x0 + v0t + ½ at2 = 36 m + (12 m/s)t + 0 = 36 m + (12 m/s)t

Step 3: You want to find the time it takes for player 1 to catch player 2 so set the equations equal to each other.

X(1) = X(2)

(2.0 m/s2)t2 = 36 m + (12 m/s)t

2.0t2 – 12t – 36 = 0

You need to solve this using the quadratic equation with a = 2.0, b = -12, and c = -36

Obviously, the correct answer will be the positive value t = 8.2 s.

(b) How far has he traveled in this time?

SOLUTION:
 

KNOWNS

UNKNOWN

v0 = 0

x

a = 4.0 m/s2

 

t = 8.2s

 

x0 = 0

 

x = x0 + v0t + ½ at2 = 0 + 0 + ½ (4.0 m/s2)(8.2 s)2

x = 134 m