Q#27
(II) A car slows down uniformly from a speed of 28 m/s to rest in 8.00 s. How far did it travel in that time?
Answer:
The words “slows down uniformly” imply that the car has a constant acceleration. The distance of travel is found from combining Eqs
$x-x_0=\overline v \Delta t$
with $\overline v=\frac{v_0+v}{2}$, so
$x-x_0=\left(\frac{v_0+v}{2}\right)\Delta t$
$x-x_0=\left(\frac{28.0 \ m/s+0 \ m/s}{2}\right)(8.00 \ s)= 112 \ m$
Q#28
(II) In coming to a stop, a car leaves skid marks 65 m long on the highway. Assuming a deceleration of 4.00 $m/s^2$ estimate the speed of the car just before braking.
The final velocity of the car is zero. The initial velocity is found from Eq.
$v^2=v_0^2+2a(x-x_0)$
with v = 0 and solving for $v_0$. Note that the acceleration is negative.
$v_0=\sqrt{v^2-2a(x-x_0)}$
$v_0=\sqrt{0^2-2(-4.00 \ m/s^2)(65 \ m)}=23 \ m/s$
Q#29
(II) A car traveling 75 km/h slows down at a constant 0.5 $m/s^2$ just by “letting up on the gas.” Calculate
(a) the distance the car coasts before it stops,
(b) the time it takes to stop, and
(c) the distance it travels during the first and fifth seconds.
Answer:
(a) The final velocity of the car is 0.
The distance is found from Eq.
$v^2=v_0^2+2a(x-x_0)$
with an acceleration of a = −0.50 $m/s^2$ and an initial velocity of 75 km/h.
$v_0=(75 \ km/h)\left(\frac{1000 \ m/km}{3600 \ s/h}\right)=20.83 \ m/s$
$x-x_0=\frac{v^2-v_0^2}{2a}$
$x-x_0=\frac{(0)^2-(20.83 \ m/s)^2}{2(-0.5 \ m/s^2)}= 434 \ m$
(b) The time to stop is found from Eq.
$v=v_0+at$
$t=\frac{v-v_0}{a}=\frac{0-20.83 \ m/s}{-0.5 \ m/s^2}=41.67 \ s$
(a) Take $x_0=x(t=0)=0$. Use Eq.
$x-x_0=v_0t-\frac{1}{2}at^2$
with a = −0.50 $m/s^2$ and an initial velocity of 75 km/h. The first second is from t = 0 s to t = 1 s, and the fifth second is from t = 4 s to t = 5 s.
$x(1)=0+(20.83 \ m/s)(1 \ s)-\frac{1}{2}(-0.5 \ m/s^2)(1 \ s)^2=20.58 \ m$
so, $x-x_0=20.58 \ m$
$x(4)=0+(20.83 \ m/s)(4 \ s)-\frac{1}{2}(-0.5 \ m/s^2)(4 \ s)^2=79.33 \ m$
$x(5)=0+(20.83 \ m/s)(5 \ s)-\frac{1}{2}(-0.5 \ m/s^2)(5 \ s)^2=97.92 \ m$
so, $x(5)-x(4)=97.92 \ m -79.33 \ m = 18.59 \ m$
Q#30
(II) Determine the stopping distances for an automobile going a constant initial speed of 95 km/h and human reaction time of 0.40 s:
(a) for an acceleration $-3.0 \ m/s^2$
(b) for $-6.0 \ m/s^2$
Answer:
The origin is the location of the car at the beginning of the reaction time. The initial speed of the car is
$v_0=(95 \ km/h)\left(\frac{1000 \ m/km}{3600 \ s/h}\right)=26.39 \ m/s$
The location where the brakes are applied is found from the equation for motion at constant velocity
$x_0=v_0t_R=(26.39 \ m/s)(0.40 \ s) = 10.56 \ m$
This is now the starting location for the application of the brakes. In each case, the final speed is 0.
(a) Solve Eq. $v^2=v_0^2+2a(x-x_0)$ for the final location.
$x=x_0+\frac{v^2-v_0^2}{2a}$
$x=10.56 \ m+\frac{0^2-(26.39 \ m/s^2)^2}{2(-3.0 \ m/s^2)}=126.6 \ m$
(b) for the final location with the second acceleration.
$x=10.56 \ m+\frac{0^2-(26.39 \ m/s^2)^2}{2(-6.0 \ m/s^2)}=69 \ m$
Q#31
(II) A driver is traveling 18.0 m/s when she sees a red light ahead. Her car is capable of decelerating at a rate of 3.65 $m/s^2$. If it takes her 0.350 s to get the brakes on and she is 20.0 m from the intersection when she sees the light, will she be able to stop in time? How far from the beginning of the intersection will she be, and in what direction?
Answer:
Calculate the distance that the car travels during the reaction time and the deceleration.
$\Delta x_1=v_0\Delta t$
$\Delta x_1=(18.0 \ m/s)(0.350 \ s)=6.3 \ m$
from eq. $v^2=v_0^2+2a\Delta x_2$,
$\Delta x_2=\frac{v^2-v_0^2}{2a}=\frac{0-(18 \ m/s)^2}{2(-3.65 \ m/s^2)}=44.4 \ m$, so
$\Delta x = 44.4 \ m + 6.3 \ m = 50.7 \ m$
Since she is only 20.0 m from the intersection, she will NOT be able to stop in time. She will be 30.7 m past the intersection.
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