If I throw a ball straight up into the air, we say the ball is an object in
"free fall"
☐ on its way up
☐ on its way back down
☑ both on its way up and on its way back down.
A ball tossed vertically upward rises, reaches its highest point, and
then falls back to its starting point. During this time, the acceleration
of the ball is always
☐ in the direction of motion
☐ opposite its velocity
☑ directed downward
☐ directed upward
I throw a ball across an open
field. At what part of its path does the ball have a minimum speed?
☐ right before it hits the ground
☐ halfway to the top
☑ at the top of its path
☐ right after it leaves my hand
☐ There's not enough information to say.
Ralph asked me a question the other day.
Consider a car accelerating forward. Its acceleration is 1.8
m/s2. During the first second, the car accelerates from 0
to 1.8 m/s. Ralph thought that since the velocity at the end of the
first second is 1.8 m/s, the car would travel 1.8 m during that first
second. But someone told him that the answer is actually 0.9 m.
Can you help Ralph understand why?
Don't just say, "Because the formula in the book says
so."
1.8 m/s is the speed of the car at the *end* of the first second. Since the car was accelerating, during the time before that, the car was at a lower speed. In fact, since it started from a standstill and the speed increases linearly (because the acceleration is constant), the average speed must be half of the final speed: 0.9 m/s.