ENPH 131 Assignment #2Solutions

Tutorial Problem (Rocket Height)

A rocket, initially at rest on the ground, accelerates straight upward with a constant acceleration of 32.2 m�s2. The rocket

accelerates for a period of 9.00 s before exhausting its fuel. The rocket continues its acent until its motion is halted bygravity. The rocket then enters free fall. Find the maximum height, ymax, reached by the rocket. Ignore air resistance and

assume a constant acceleration due to gravity of 9.81 m�s2.

For consistency in labeling, let us define y º s. During the first portion of the trip (first 9s), while the rocket is being pro-

pelled, vo = 0, a = 39.2 m�s2, yo = 0. Apply eqn. 12-5:

y = yo + vo t +1

2a t2

y1 = 0 + 0 +1

2H39.2L I92M

y1 = 1587.6 m

Also, use eqn. 12-4 to calculate the velocity of the rocket at the time the fuel runs out:

v = vo + a tv1 = 0 + 39.2 H9Lv1 = 352.9 m �s

For the second portion of the trip, from the time the fuels runs out until the rocket reaches its maximum heigh (when v = 0),

yo = y1 = 1587.6 m, vo = v1 = 352.8 m �s, v = 0, a = -9.81 m�s2. Apply eqn. 12-6:

v2 = vo2 + 2 aHy - yoL

0 = 352.82 + 2 H-9.81L Hymax - 1587.6Lymax = 7931.5 m > 7930 m

Problem 12.32Ball A is thrown vertically upward from the top of a 25 m-high-building with an initial velocity of 7 m �s. At the sameinstant another ball, B, is thrown upward from the ground with an initial velocity of 19 m �s.

a) Determine the height from the ground at which they pass each other.

Take the ground to be s = 0.

Ball A: so = 25 m, vo = 7 m �s, a = -9.81 m �s. Apply eqn. 12-5:

s = so + vo t +1

2a t2

sA = 25 + 7 t +1

2H-9.81L t2

s = so + vo t +1

2a t2

sA = 25 + 7 t +1

2H-9.81L t2

Ball B: so = 0, vo = 19 m �s, a = -9.81 m�s2. Apply eqn. 12-5:

s = so + vo t +1

2a t2

sB = 0 + 19 t +1

2H-9.81L t2

When the balls pass each other:

sA = sB

25 + 7 t +1

2H-9.81L t2 = 19 t +

1

2H-9.81L t2

12 t = 25t = 2.08 s

To obtain the height when the balls pass each other, plug t into the eqn. for sB:

s = sB = 19 H2.08L +1

2H-9.81L I2.082M

s = 18.3 m

b) Determine the time at which they pass each other.

As shown above, the balls pass each other at t = 2.08 s.

Problem 12.33

A motorcycle starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft�s2 until it reaches

a speed of 50 ft �s. Afterwards it maintains this speed. Also, when t = 0, a car located 6000 ft down the road is travelingtoward the motorcycle at a constant speed of 60 ft �s.

a) Determine the time when they pass each other.

For the first portion of the trip, as the motorcycle accelerates:

Motorcycle: vo = 0, v = 50 ft �s, a = 6 ft�s2. Apply eqn. 12-4:

v = vo + a t50 = 0 + 6 t1t1 = 8.333 s

Thus, the accelerating portion of the trip takes t1 = 8.333 s. At this time, t1, let us calculate the position of the motorcycleand car using eqn. 12-5:

Motorcycle:

s = so + vo t +1

2a t2

sm = 0 + 0 +1

2H6.0L I8.3332M

sm = 208.3 m

Car:

s = so + vo t +1

2a t2

sc = 6000 + -60 H8.333L + H0Lsc = 5500 m

2 ENPH 131 assignment 2 solutions.nb

s = so + vo t +1

2a t2

sc = 6000 + -60 H8.333L + H0Lsc = 5500 m

After t = 8.333 s both vehicles are travelling with constant velocities: vm = 50 ft �s, vc = 60 ft �s. Use eqn. 12-5 to determinethe time, t2, when they pass (Note: t2 is the time after t1, the initial acceleration period):

Motorcycle:

s = so + vo t +1

2a t2

sm2 = 208.3 + 50 t + 0

Car:

s = so + vo t +1

2a t2

sc2 = 5500 - 60 t + 0

When the vehicles pass each other sm2 = sc2:

sm2 = sc2

208.3 + 50 t2 = 5500 - 60 t2t2 = 48.11 s

Thus, from t = 0, the time when the vehicles pass each other is

t = t1 + t2t = 8.333 + 48.11t = 56.4 s

b) Determine the distance travelled by the motorcycle when they pass each other.

Plug t2 into the eqn. for sm2:

sm2 = 208.3 + 50 tsm2 = 208.3 + 50 H48.11Lsm2 = 2614 m > 2610 m

Tutorial Problem (Graph of Sports Car’s Velocity)As shown, the velocity v of a sports car is graphed as a function of time t.

ENPH 131 assignment 2 solutions.nb 3

a) Find the sports car’s maximum velocity, vmax, during the 10s interval depicted in the graph.

From the graph above we see that the car’s maximum velocity occurs at t = 4 s and is roughly 55 m �s.

vmax = 55 m � s

b) During which time interval is the acceleration positive?

Acceleration is given by the derivative of velocity with respect to time. On the graph avobe, this is the slope. We see thatfrom t = 0 ® t = 4 s the slope of the curve is positive, where as from t = 4 s ® t = 10 s the slope is negetive. Therefore thetime interval during which acceleration is positive is

a > 0 for t = 0 ® t = 4 s

c) Find amax, the magnitude of the sports car’s maximum acceleration.

The maximum acceleration (in magnitude) will occur at the section of the graph with the steepest slope (whether positive ornegetive). We can see this occurs from t = 0 ® t = 1 s when the car accelerates from v = 0 ® to v = 30 m �s:

amax =v-vo

t-to=

30

1

amax = 30 m � s2

d) Find amin, the minimum magnitude of the sports car’s acceleration.

The minimum acceleration (in magnitude) will occur at the section of the graph with the shallowest slope. We can see thisoccurs at t = 4 s where, in fact, slope = 0. Therefore

amin = 0.0 m � s2

e) Find the distance r0,2 traveled by the car between t = 0 s and t = 2 s.

From t = 0 s ® t = 1 s, vo = 30 m �s, a = 30 m�s2. Apply eqn. 12-5:

s = so + vo t +1

2a t2

s1 = 0 + 0 +1

2H30L I12M

s1 = 15 m

4 ENPH 131 assignment 2 solutions.nb

s = so + vo t +1

2a t2

s1 = 0 + 0 +1

2H30L I12M

s1 = 15 m

From t = 1 s ® t = 2 s, so = s1 = 15 m, vo = v1 = 30 m �s, a = 20 m�s2. Apply eqn. 12-5:

s = so + vo t +1

2a t2

s2 = 15 + 30 +1

2H20L I12M

s2 = 55 m

Problem 12.60A motorcyclist starting from rest travels along a straight road and, for 10s, has an acceleration as shown below.

a) Find the distance traveled in the 10s.

For the first 6s, a =1

6t2. Use eqn. 12-2 to obtain an expression for the motorcycle’s velocity:

a â t = â v

Ùt

0

1

6t2 â t = Ù

0

v

â v

v =1

18t3

Now use eqn. 12-1 to obtain an expression for the motorcycle’s displacement:

v â t = â s

Ù0

t1

18t3 â t = Ù

s

0

â s

s =1

72t4

After 6s:

s6 =1

72I64M

s6 = 18 m

ENPH 131 assignment 2 solutions.nb 5

s6 =1

72I64M

s6 = 18 m

and

v6 =1

18I63M

v6 = 12 m �s

For the next 4s the motor cycle has a constant acceleration, a = 6 m�s2. Apply eqn. 12-5:

s = s6 + v6 t +1

2a t2

s = 18 + 12 H4L +1

2H6L I42M

s = 114 m

b) Draw the v - t that describes the motion.

As shown above the velocity of the motorcycle for the first 6s is

v =1

18t3

After that it is given by

v = vo + a t ' = v6 + a Ht - t6Lv = 12 + 6 Ht - 6L

Using these two equations we can compile a table for the motorcycle’s velocity as a function of time:

Time HsL Velocity Hm �sL

0 0

1 0.06 > 0.1

2 0.4

3 1.5

4 3.6

5 6.9

6 12.0

7 18.0

8 24.0

9 30.0

10 36.0

6 ENPH 131 assignment 2 solutions.nb

ENPH 131 assignment 2 solutions.nb 7