Q#17
(I) A sports car accelerates from rest to 95 km/h in 4.3 s. What is its average acceleration in $m/s^2$?
Answer:
The average acceleration is found from Eq.
$a=\frac{\Delta v}{\Delta t}$
$a=\frac{95 \ km/h-0}{4.3 \ s}$
$a = 22.09 \ km/h/s$ or
$a = (22.09 \ km/h/s)\left(\frac{1000 \ m}{1 \ km}\right)\left(\frac{1 \ h}{3600 \ s}\right)$
$a=6.1 \ m/s^2$
Q#18
(I) A sprinter accelerates from rest to 9.00 m/s in 1.38 s. What is her acceleration in
(a) $m/s^2$
(b) $km/h^2$
Answer:
(a) The average acceleration of the sprinter is
$\overline a=\frac{\Delta v}{\Delta t}$
$a=\frac{9.00 \ m/s-0.00 \ m/s}{1.38 \ s}=6.52 \ m/s^2$
(b) We change the units for the acceleration
$a = (6.52 \ m/s^2)\left(\frac{1 \ km}{1000 \ m}\right)\left(\frac{3600 \ s}{1 \ h}\right)^2$
$a=8.45 \times 10^4 \ km/h^2$
Q19
(II) A sports car moving at constant velocity travels 120 m in 5.0 s. If it then brakes and comes to a stop in 4.0 s, what is the magnitude of its acceleration (assumed constant) in and in $m/s^2$ g’s A(g = 9.80 $m/s^2$)?
Answer:
The initial velocity of the car is the average velocity of the car before it accelerates.
$\overline v=\frac{\Delta x}{\Delta t}$
$\overline v=\frac{120 \ m}{5.0 \ s}=24 \ m/s=v_0$
The final velocity is υ = 0, and the time to stop is 4.0 s. Use Eq. $v=v_0+at$ to find the acceleration.
$a=\frac{v-v_0}{t}$
$a=\frac{0-24 \ m/s}{4.0 \ s}=-6.0 \ m/s^2$
Thus the magnitude of the acceleration is $6.0 \ m/s^2$ or
$a = (6.0 \ m/s^2)\left(\frac{g}{9.8 \ m/s^2}\right)=0.61$ g's
Q#20
(II) At highway speeds, a particular automobile is capable of an acceleration of about 1.8 $m/s^2$. At this rate, how long does it take to accelerate from 65 km/h to 120 km/h?
Answer:
Given: acceleration is a = 1.8 $m/s^2$
initial velocity, $v_i=65 \ km/h = (65 \ km/h)\left(\frac{1000 \ m/km}{3600 \ s/h}\right)=18.06 \ m/s^2$
final velocity, $v_i=120 \ km/h = (120 \ km/h)\left(\frac{1000 \ m/km}{3600 \ s/h}\right)=33.33 \ m/s^2$
We assume that the speedometer can read to the nearest km/h, so the value of 120 km/h has three significant digits.
The time can be found from the average acceleration,
$\overline a=\frac{\Delta v}{\Delta t}$
$\Delta t=\frac{\Delta v}{\overline a}$
$\Delta t=\frac{33.33 \ m/s-18.06 \ m/s}{1.8 \ m/s^2}$
$\Delta t=8.488 \ s \approx 8.5 \ s$
Q#21
(II) A car moving in a straight line starts at x = 0 at t = 0. It passes the point x = 25.0 m with a speed of 11.0 m/s at 3.00 s. It passes the point with a speed x = 385 m of 45.0 m/s at t = 20.0 s. Find
(a) the average velocity, and
(b) the average acceleration, between and 3.00 s and 20.0 s.
Answer:
(a) the average velocity is
$\overline v=\frac{\Delta x}{\Delta t}$
$\overline v=\frac{385 \ m - 25.0 \ m}{20.0 \ s - 3.0 \ s}=21.2 \ m/s$
(b) the average acceleration,
$\overline a=\frac{\Delta v}{\Delta t}$
$\overline a=\frac{45.0 \ m/s - 11.0 \ m/s}{20.0 \ s - 3.0 \ s}=2.00 \ m/s^2$
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