Question

. In this class, we will use the value of g-10 m/s2g-10 m/s2, or g-9.8 m/s2g-9.8 m/s2, when more precision is needed. According to your textbook, what is the typical variation of gravitational acceleration gg on the surface of the Earth? The value of gg ranges from m/s2m/s2 to m/s2m/s2 . A bowling ball rolls along a straight track with an average velocity of 11 cm/s for 20 seconds. How far does the bowling ball roll in this time? The bowling ball rollsmeters . A jogger running at 5.5 m/s5.5 m/s starts running up a small hill, slowing down at an average rate of 0.2 m/s20.2 m/s2 for the next 3 seconds while going to the top of the hill. As they reach the top of the hill, how fast is the jogger running? (Or "running", depending on how much they slowed down.) The joggers final speed ism/sm/s. . If your reaction time is 0.18 second, in the time it takes for you to react (and grab the object, etc.), how far does an object drop, when dropped from rest? Use the approximate value g-10 m/s2g-10 m/s2 in your calculation. The object drops cm in 0.18 second.

Answer 1

The value of g ranges from $9.780\,m/s^2$ to $9.832\,m/s^2$

!)

Average velocity = (total distance )/(total time)

$v_{avg}=\frac{d}{t}$

$d=v_{avg}t=11*20=220\,cm=2.20\,m$

The bowling ball rolls $2.20\,m$

2)

Initial velocity of jogger $u=5.5\,m/s$

He is slowing with an acceleration $a=-\frac{0.2}{3}=-0.067\,m/s^2$

Final velocity of jogger is $v=u+at=5.5-0.067*3=5.3\,m/s$

3)

Reaction time $t=0.18\,s$ , object is dropped from rest, $u=0\,m/s$

Using $S=ut+\frac{1}{2}at^2$

$a=g=10\,m/s^2$

$S=0+\frac{1}{2}(10)(0.18)^2=0.162\,m=16.2\,cm$

The object drops in $16.2\,cm$ in $t=0.18\,s$