Q#22
(I) A car slows down from 28 m/s to rest in a distance of 88 m. What was its acceleration, assumed constant?
Answer:
The acceleration can be found from Eq.
$v^2=v_0^2+2a(x-x_0)$
$a=\frac{v^2-v_0^2}{2(x-x_0)}$
$a=\frac{0^2-(28 \ m/s)^2}{2(88 \ m)}=-4.5 \ m/s^2$
Q#23.
(I) A car accelerates from 14 m/s to 21 m/s in 6.0 s. What was its acceleration? How far did it travel in this time? Assume constant acceleration.
Answer:
By definition, the acceleration is
$a = \frac{v-v_0}{t}=\frac{21 \ m/s-14 \ m/s}{6.0 \ m/s}= 1.167 \ m/s^2$
The distance of travel can be found from Eq.
$x-x_0=v_0t+\frac{1}{2}at^2$
$x-x_0=(14 \ m/s)(6.0 \ s)+\frac{1}{2}(1.167 \ m/s^2)(6.0 \ s)^2=105 \ m$
It can also be found from Eq
$x-x_0=\overline v \Delta t$
with $\overline v=\frac{v_0+v}{2}$, so
$x-x_0=\left(\frac{v_0+v}{2}\right)\Delta t$
$x-x_0=\left(\frac{14 \ m/s+21 \ m/s}{2}\right)(6.0 \ s)= 105 \ m$
Q#24.
(I) A light plane must reach a speed of 35 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 $m/s^2$.
Answer:
Assume that the plane starts from rest. The distance is found by solving Eq.
$v^2=v_0^2+2a(x-x_0)$
$x-x_0=\frac{v^2-v_0^2}{2a}$
$x-x_0=\frac{(35 \ m/s)^2-0^2}{2(3 \ m/s^2)}= 204.2 \ m$
Q#25.
(II) A baseball pitcher throws a baseball with a speed of Estimate the average acceleration of 43 m/s the ball during the throwing motion. In throwing the baseball, the pitcher accelerates it through a displacement of about 3.5 m, from behind the body to the point where it is released (Fig. 2–37)
Q#26.
(II) A world-class sprinter can reach a top speed (of about 11.5 m/s) in the first 18.0 m of a race. What is the average acceleration of this sprinter and how long does it take her to reach that speed?
Answer:
The sprinter starts from rest. The average acceleration is found from Eq.
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