Chapter 6: Q.78 (page 158)

An object with cross section A is shot horizontally across frictionless ice. Its initial velocity is $v_{0 x}$ at $t_{0} = 0 s$ . Air resistance is not negligible.
a. Show that the velocity at time t is given by the expression
$v_{x} = \frac{v_{0 x}}{1 + C ρ A v_{0 x} t / 2 m}$
b. A $1.6 - m$ -wide, $1.4 - m$ -high, $1500 k g$ car with a drag coefficient of $0.35$ hits a very slick patch of ice while going $20 m / s$ . If friction is neglected, how long will it take until the car’s speed drops to $10 m / s$ ? To $5 m / s$ ?
c. Assess whether or not it is reasonable to neglect kinetic friction.

Short Answer

Expert verified

a. derivation of the velocity at time $t .$

b. time taken for car to drop its speed is $312.5 s$

c. Its absurd. so that kinetic friction were neglected.

Step by step solution

Given information (part a)

cross section of the object $A$

Initial velocity $v_{0 x}$ at $t = 0 s$

velocity at time $t = v_{x}$

Coefficient of drag $C$

density of air $ρ$

mass of the object $m$

Explanation (part a)

From Newton's second law

$F = m a$

The drag force is

$D = \frac{1}{2} C ρ A (v_{x})^{2}$

Consider the negative sign for drag force as it acts opposite to the net force and equal the drag force.

$m a = \frac{1}{2} C ρ A (v_{x})^{2} a = \frac{C ρ A}{2 m} (v_{x})^{2}$

Here $a = \frac{d v_{x}}{d t}$

Therefore, we get role="math" localid="1650212860312" $\frac{d v_{x}}{d t} = \frac{C ρ A}{2 m} (v_{x})^{2} \frac{d v_{x}}{(v_{x})^{2}} = \frac{C ρ A}{2 m} d t$

On integrating the above equation, we get

$\int_{v_{0 x}}^{v_{x}} \frac{d v_{x}}{(v_{x})^{2}} = \frac{C ρ A}{2 m} \int_{0}^{t} d t {(\frac{- 1}{(v_{x})})}_{v_{0 x}}^{v_{x}} = \frac{C ρ A}{2 m} {(t)}_{0}^{t}$

$v_{x} = \frac{v_{0 x}}{1 + C ρ A v_{0 x} t / 2 m}$

Conclusion: The expression for velocity at time $t$ has been derived.

Given information (part b)

length of the car $l = 1.6 m$

height of the car $h = 1.4 m$

mass of the car $m = 1500 k g$

drag coefficient $C = 0.35$

Initial velocity role="math" localid="1650213612097" $v_{0 x} = 10 m / s$

Final velocity $v_{x} = 5 m / s$

density of air $ρ = 1.225 k g / m^{3}$

Explanation (part b)

The velocity expression at time $t$ is given by

$v_{x} = \frac{v_{0 x}}{1 + C ρ A v_{0 x} t / 2 m}$

Plugging all the values in the above equation, we get

$5 = \frac{10}{1 + \frac{(0.35) (1.225) (1.6 \times 1.4) (10) t}{2 \times 1500}} 5 = \frac{10}{1 + \frac{(0.35) (1.225) (1.6 \times 1.4) (10) t}{2 \times 1500}} 5 = \frac{10}{1 + 0.0032 t} 1 + 0.0032 t = 2 t = 312.5 s e c$

The time taken for the car to drop its speed $312.5 s e c s .$

Explanation (part c)

If we consider an object that moves by inertia with the drag or frictional force,the net force is not zero and the motion will be decelerated (non-uniform), and we will have to introduce new forces to make it uniform, which is absurd. That is why we neglect the mentioned forces.