1. A tank contains 20 liters of pure water. Salt water containing 40 grams of salt per liter is pumped into the tank at 4 liters per minute. (a) Express the salt concentration C(t) after t minutes (in g/L). (b) What is the long-term concentration of salt, i.e., limt∞ C(t)?

Solution:

a. The concentration is, in units of g/L,

C(t) = total salt/total volume = (40 ∙ 4 ∙ t) / (20 + 4 ∙ t) = 40t/ (5 + t)

b. The long-term concentration is, in units of g/L,

limt∞ 40t/ (5 +t) = limt∞ 40t/ (5 + t) ∙ 1/t / 1/t

= limt∞ 40/ (5/t + 1) = 40

2. A cheetah is crouched 10 m to the east of an observer’s vehicle. At time t = 0 the cheetah charges an antelope and begins to run along a straight line. During the first 2.0 s of the attack, the cheetah’s coordinate x varies with time according to the equation x = 10 m + (5.0 m/s2)t2. Derive a general expression for the instantaneous velocity as a function of time, and from it find vx at t = 1.0 s and t = 2.0 s.

Solution:

To find the instantaneous x-velocity as a function of time, take the derivative of the expression for x with respect to t. The derivative of a constant is zero, and for any n the derivative of tn is ntn-1, so the derivative of t2 is 2t. Therefore

vx = dx/dt = (5.0 m/s2)(2t) = (10 m/s2)t

At time t = 1.0 s, vx = 10 m/s2. At time t = 2.0 s, vx= 20 m/s.

3. Suppose the x –velocity of a car at any time rate is given by the equation

vx = 50 m/s + (0.50 m/s2)t2

Derive an expression for the instantaneous x-acceleration at any time, and use it to find the x-acceleration at t = 1.0 s and t = 3.0 s.

Solution:

The instantaneous x-acceleration for is ax = dvx / dt. The derivative of a constant is zero and the derivative of t2 is 2t, so

ax = dvx / dt = d/ dt [50 m/s + (0.50 m/s3)t2] = (0.50 m/s3)(2t) = (1.0 m/s3)t

When t = 1.0 s,

ax = (1.0 m/s3)(1.0 s) = 1.0 m/s3

When t = 3.0 s,

ax = (1.0 m/s3)(3.0 s) = 3.0 m/s2

4. To study damage to aircraft that collide with large birds, a person designed a test gun that will accelerate chicken-sized objects so that their displacement along the gun barrel is given by

x = (8.0 x 103 m/s2)t2 – (7.0 x 104 m/s3)t3

The object leaves the end of the barrel at t = 0.025 s.

a. How long must the gun barrel be?b. What will be the speed of the objects as they leave the end of

the barrel?

Solution:

a. x = (8.0 x 103 m/s2)t2 – (7.0 x 104 m/s3)t3

x = 0 at t = 0When t = 0.025 s, x = (8.0 x 103 m/s2)(0.025)2 – (7.0 x 104 m/s3)(0.025)3 = 3.9 m.The length of the barrel must be 3.9 m.

b. vx = dx/dt = (16.0 x 103 m/s2)t – (21.0 x 104 m/s3)t2

At t = 0, vx = 0 (object starts from rest).At t = 0.025 s, when the object reaches the end of the barrel,vx = (16.0 x 103 m/s2)(0.025 s) – (21.0 x 104 m/s3)(0.025 s)2 = 268.75 m/s

5. Refer to number 5. What net force must be exerted on a 3.00-kg object at (i) t = 0 and (ii) t = 0.025 s?

Solution:

∑ Fx = max, so must find ax.

ax = dvx/dt = 16 x 103 m/s2 – (42.0 x 104 m/s3)t

(i) At t = 0, ax = 16 x 103 m/s2 and ∑ Fx = (3.00 kg)(16 x 103 m/s2) = 4.8 x 104 N.

(ii) At t = 0.025 s, ax = 16 x 103 m/s2 – (42.0 x 104 m/s3)(0.025 s) = 6.0 x 103 m/s2 and ∑ Fx = (1.50 kg)(6.0 x 103

m/s2) = 5.5 x 103 N.

6. A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t) = bt2 – ct3, where b = 2.50 m/s2 and c = 0.150 m/s3. Calculate the average velocity of the car for the time interval t = 0 to t = 10.0 s.

Solution:

vav-x = ∆x/∆tt = 10.0 s: x = (2.50 m/s2)(10.0 s)2 – (0.150 m/s3)(10.0 s)3 = 250 m – 150 m = 100 mThen vav-x = ∆x/∆t = 100 m/ 10.0 s = 10.0 m/s.

2b = 3ct so t = 2b/3c = 2 (2.50 m/s2)/ 3(0.150 m/s3) = 11.1 s

7. Refer to number 7. a. Calculate the instantaneous velocity of the car at t = 0, t =

5.0 s, and t = 10.0 s. b. How long after starting from rest is the car again at rest?

Solution:

a. vx = dx/dt = 2bt – 3ct2

i. t = 0, vx = 0ii. t = 5.0 s: vx = 2 (2.50 m/s2)(5.0 s) – 3(0.150 m/s3)(5.0

s)2 = 25.0 m/s – 11.25 m/s = 13.75 m/s.iii. t = 10.0 s: vx = 2 (2.50 m/s2)(10.0 s) – 3(0.150 m/s3)

(10.0 s)2 = 50.0 m/s – 45.0 m/s = 5.0 m/s.

b. vx = 2bt – 3ct2

vx = 0 at t = 0.vx = 0 next when 2bt – 3ct2 = 0

8. A car’s velocity as a function of time is given by vx(t) = α + βt2, where α = 3.00 m/s and β = 0.150 m/s3.

a. Calculate the average acceleration for the time interval t = 0 to t = 5.00s.

b. Calculate the instantaneous acceleration for t = 0 and t = 5.00s.

Solution:

The average acceleration is aav-x = ∆vx/∆t. The instantaneous acceleration is ax = dvx/dt.

vx(0) = 3.00 m/s and vx(5.00 s) = 6.75 m/s.

a. aav-x = ∆vx/∆t = (6.75 m/s – 3.00 m/s)/ 5.00 s = 0.750 m/s2

b. ax = dvx/dt = (0.150 m/s3)(2t) = (0.300 m/s3)t. At t = 0, ax = 0. At t = 5.00 s, ax = 1.50 m/s2.

9. The position of the front bumper of a test car under micro-processor control is given by x(t) = 2.15 m + (4.60 m/s2)t2 – (0.100 m/s6)t6. Find its position and acceleration at the instants the car has zero velocity.

Solution:

vx(t) = dx/dt and ax(t) = dvx/dtd/dt(tn) = ntn-1 for n ≥ 1.

vx(t) = (9.20 m/s2)t – (0.600 m/s6)t5 and ax = 9.20 m/s2 – (3.00 m/s6)t4. Setting vx = 0 gives t = 0 and t = 1.98 s. At t = 0, x = 2.15 m and ax = 9.20 m/s2. At t = 1.98 s, x = 14.16 m and ax = - 171.56 m/s2.

10. A particle moves along the x axis according to the equation x = 4.00 + 6.00t – 2.00t2, where x is in meters and t is in seconds. At t = 3.00 s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration.

Solution:

x = 4.00 + 6.00t – 2.00t2, v = dx/dt = 6.00 – 4.00t, a = dv/dt = –4.00

At t = 3.00 s: a. x = (4.00 + 18.00 – 6.00)m = 16.00 mb. v = (6.00 – 12.00) m/s = - 6.00 m/sc. a = –4.00 m/s2

11. The position of a particle moving along the x-axis varies in time according to the expression x = 6t2, where x is in meters and t is in seconds. Evaluate its position (a) at t = 6.00 s and (b) at 6.00 s + ∆t. (c) Evaluate the limit of ∆x/∆t as ∆t approaches zero, to find the velocity at t = 6.00 s.

Solution:

a. At any time, t, the position is given by x = (6.00 m/s2)t2. Thus, at ti = 6.00 s: xi = (6.00 m/s2)(6.00 s)2 = 216.0 m

b. At tf = 6.00 s + ∆t: xf = (6.00 m/s2)(6.00 s + ∆t)2, or

xf = 216.0 m + (72.0 m/s)∆t + (6.00 m/s2)(∆t)2

c. The instantaneous velocity at t = 6.00 s is:

v = lim∆t 0 (xf – xi/ ∆t) = lim∆t 0 (72.0 m/s + (12.00 m/s2)∆t) = 72.0 m/s.

12. The velocity at any time t is vxi = (50 – 5t2) m/s and the velocity at any time t + ∆t is

vxf = 50 – 5(t + ∆t)2 = 50 – 5t2 – 10t∆t – 5(∆t)2

Therefore, the change in velocity over the time interval ∆t is

∆vx = vxf – vxi = [ – 10t ∆t – 5(∆t)2] m/s

Dividing this expression by ∆t and taking the limit of the result as ∆t and taking the limit of the result as ∆t approaches zero gives the acceleration at any time t:

ax = lim ∆t 0 ∆vx/∆t = lim ∆t 0 ( – 10t – 5∆ t) = –10t m/s2

Therefore, at t = 2.0s,

ax = (–10)(2.0) m/s2 = –20 m/s2

Because the velocity of the particle is positive and the acceleration is negative, the particle is slowing down.

13. Two bicycles leave an intersection at the same time. One heads north going 24 mph and the other heads east going 10 mph. How fast are the bikes getting away from each other after one hour?

Solution:

Set up a coordinate system with the origin at the intersection and the -axis pointing north. We assume that the position of the bike heading north is a function of the position of the bike heading east.

y = (24/10) x

The distance between the bikes is given by

s =√(x2 + y2) = √{(x2 + [(24/10)x]2} = √{[576 + 100/100]x2} = √[(676/100)x2] = (13/5)x

Let t represent the elapsed time in hours. Find ds/dt when t = 1. Apply the chain rule to s:

ds/dt = (ds/dx)(dx/dt) = (13/5) ∙ 5 = 13

Thus, the bikes are moving away from one another at 13 mph.

14. Suppose s(t) = 3t 2 represents the position of a race car along a straight track, measured in feet from the starting line at time t seconds. (a) What is the average rate of change of s(t) from t = 3 to t = 4? (b) What is the instantaneous rate of change of the same race car at time t = 2?

Solution:

a. The average rate of change is equal to the total change in position divided by the total change in time:

Average Rate = ∆s/∆t = [s(4) – s(3)] / [4 – 3] = (48 – 27) / 1 =21 ft/s

The average velocity of the car between times t = 2 and t = 3 is 21 feet per second.

b. The instantaneous rate of change measures the rate of change, or slope, of a curve at a certain instant. Thus, the instantaneous rate of change is given by the derivative. In this case, the instantaneous rate is s'(2) .

s'(t)=6t  

s'(2)=6(2) = 12 feet per second  

Thus, the derivative shows that the racecar had an instantaneous velocity of 12 feet per second at time t = 2.

15. Suppose the Helium gas was pumped into a hot air balloon at a rate of 3 m3/min. How fast is the radius of the cell increasing when the radius is 25 m?

Solution:

V= 43π r3→dV

dt (V=43π r3)→dV

dt=4 π r2 dr

dt

Substituting dVdt =3m3/min and r=25m into the last equation:

3=4 π (25)2 drdt

drdt

=3.82 x 10hhhh− 4 meters per minute

16. Alex is travelling from school to Johann’s house with her scooter according to the equation: d(t) = 10 – 3t2. What is her speed at t = 5 s of her trip? In this instance, space is measured in meters and time in seconds.

Solution:

v’5 = lim 10 – 3(5 + Δ t ) 2 – (10 – 3 ∙ 5 2 ) = lim -15 Δ t – 3( Δ t ) 2 Δt -> 0 Δt Δt -> 0 Δtlim Δ t (-15 -3 Δ t ) = (-15 - 0) = -15 m/s

Δt -> 0 Δt

17. Alex placed a 20 foot ladder against a wall smeared with marmalade. The ladder slips away from the wall at a rate of 0.02 ft/sec. How fast is the top of the ladder sliding down the wall when the top is 16 feet above the ground?

Solution:

x2+ y2=202→ ddt

(x2+ y2=400 )→2x dxdt

+2 y dydt

=0

When y is 16 ft: x2+162=202→x=12 ft .

2 (12 ) (0.02 )+2 (16 ) dydt

=0

dydt

=−0.015 feet per second

The negative sign indicates that y is decreasing.

18. One morning, Alex who is 5’4 ft tall, walks away from a flagpole 20 feet high at a rate of 5 ft /h. How fast is the end of her shadow moving?

Solution:

Let x be the distance of Jinlian from the lamp post and y be the distance from the tip of her shadow to the lamp post. By similar triangles,

205.417

= yy−x

→20 y−20 x=5.417 y→14.583 y=20 x

ddt

(14.583 y=20 x )→14.583 dydt

=20 dxdt

14.583 dydt

=20 (5 )

dydt

=0.15 ft /h

19. A conical water tank with vertex down has a radius of 5.5m at the top and is 10m high. If water flows out of the tank at a rate of 1m3 /min, how fast is the depth of the water decreasing when the depth is 5 m deep?

Solution:

V=13π r2h

By similar triangles, we can say that 5.510

= rh→r=11

20h

V= 13π ( 11

20h)

2

h→V=0.576h3→ ddt

(V=0.576h3 )→dVdt

=0.576h2 d hdt

Substituting dVdt =−1m3/min¿h=5 :

−1=0.576(5)2 d hdt

d hdt

=−0.069meters per minute

The negative sign indicates that y is decreasing.

20. A box is to be made by cutting out squares of the same size from the corners of a rectangular piece of cardboard 21 in. by 23 in. and turning up the sides. Prove that expressing the volume of the plate as a function of the length of the side of the square cut out is continuous on its domain.

Solution:

v=lwh= (21−2x ) (23−2x ) x ;where 0≤x ≤ 187

The function is continuous on its domain because v is a polynomial.