Q#68
A car is behind a truck going 18.0 m/s on the highway. The car’s driver looks for an opportunity to pass, guessing that his car can accelerate at 0.60 $m/s^2$ and that he has to cover the 20-m length of the truck, plus 10-m extra space at the rear of the truck and 10 m more at the front of it. In the oncoming lane, he sees a car approaching, probably at the speed limit, 25 m/s (55 mph). He estimates that the car is about 500 m away. Should he attempt the pass? Give details.
Answer:
The critical condition is that the total distance covered by the passing car and the approaching car must be less than 500 m so that they do not collide.
The passing car has a total displacement composed of several individual parts. These are
(i) the 10 m of clear room at the rear of the truck,
(ii) the 20-m length of the truck,
(iii) the 10 m of clear room at the front of the truck, and
(iv) the distance the truck travels. Since the truck travels at a speed of $\overline{v}$ = 18 m/s, the truck will have a displacement of
$\Delta x_{truck}=\overline{v}t=(18 \ m/s)t$.
Thus the total displacement of the car during passing is
$\Delta x_{passing \ car}=40 \ m+(18 \ m/s)t$.
To express the motion of the car, we choose the origin to be at the location of the passing car when the decision to pass is made.
For the passing car, we have an initial velocity of $v_0$ = 18 m/s and an acceleration of a = 1.0 $m/s^2$.
Find $\Delta x_{passing \ car}$ from Eq. $x-x_0=v_0t+\frac{1}{2}at^2$.
$\Delta x_{passing \ car}=x-x_0=v_0t+\frac{1}{2}at^2$
$\Delta x_{passing \ car}=(18 \ m/s)t+\frac{1}{2}(0.60 \ m/s^2)t^2$
Set the two expressions for $\Delta x_{passing \ car}$ equal to each other in order to find the time required to pass.
$40 \ m + (18.0 \ m/s)t_{pass}=(18.0 \ m/s)t_{pass}+\frac{1}{2}(0.60 \ m/s^2)t_{pass}^2$
$40 \ m =\frac{1}{2}(0.60 \ m/s^2)t_{pass}^2$
$t_{pass}=\sqrt{\frac{80 \ s^2}{0.6}}=11.55 \ s$
Calculate the displacements of the two cars during this time.
$\Delta x_{passing \ car}$ = 40 m + (18 m/s)(11.55 s) = 247.9 m
$\Delta x_{approaching \ car}=v_{approaching \ car}t$
$\Delta x_{approaching \ car}$ = (25 m/s)(11.55 s) = 288.75 m
Thus the two cars together have covered a total distance of 247.9 m + 288.75 m = 536.65 m, which is more than allowed. The car should not pass.
Q#69
Agent Bond is standing on a bridge, 15 m above the road below, and his pursuers are getting too close for comfort. He spots a flatbed truck approaching at 25 m/s, which he measures by knowing that the telephone poles the truck is passing are 25 m apart in this region. The roof of the truck is 3.5 m above the road, and Bond quickly calculates how many poles away the truck should be when he drops down from the bridge onto the truck, making his getaway. How many poles is it?
Answer:
Choose downward to be the positive direction and $y_0$ = 0 to be at the height of the bridge.
Agent Bond has an initial velocity of $v_{0y}$ = 0, an acceleration of a = g, and will have a displacement of y = 15 m − 3.5 m = 11.5 m.
Find the time of fall from Eq. $y-y_0=v_{0y}t+\frac{1}{2}a_yt^2$ with x replaced by y.
$y-0=0+\frac{1}{2}a_yt^2$
$t = \sqrt{\frac{2y}{a_y}}$
$t = \sqrt{\frac{2(11.5 \ m)}{9.80 \ m/s^2}}=1.532 \ s$
If the truck is approaching with v = 25 m/s, then he needs to jump when the truck is a distance away given by d = vt = (25 m/s)(1.532 s) = 38.3 m. Convert this distance into “poles.”
d = (38.3 m)(1 pole/25 m) = 1.53 pole
So he should jump when the truck is about 1.5 poles away from the bridge.
Q#70
A conveyor belt is used to send burgers through a grilling machine. If the grilling machine is 1.2 m long and the burgers require 2.8 min to cook, how fast must the conveyor belt travel? If the burgers are spaced 25 cm apart, what is the rate of burger production (in burgers/min)?
Answer:
The speed of the conveyor belt is found from Eq. $\Delta x = \overline{v}t$ for average velocity.
$\overline{v}=\frac{\Delta x}{\Delta t}$
$\overline{v}=\frac{1.2 \ m}{2.8 \ min}$ = 0.4286 m/min
The rate of burger production, assuming the spacing given is center to center, can be found as follows:
$\left(\frac{1 \ burger}{0.25 \ m}\right)\left(\frac{0.4286 \ m}{1 \ min}\right)=1.7$ burger/min
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