Physics questions... about a velocity rock and vectors and stuff.?

Question

A large boulder is ejected vertically upward from a volcano with an initial speed of 39.8 m/s. Air resistance may be ignored.\

At what time after being ejected is the boulder moving at a speed 20.4 m/s upward?

At what time is it moving at a speed 20.4 m/s downward?

When is the displacement of the boulder from its initial position zero?

When is the velocity of the boulder zero?

What is the magnitude of the acceleration while the boulder is moving?

What is the direction of the acceleration while the boulder is moving?

I just have no clue how to do these... I keep getting the first one wrong and it throws off everything. It sucks.

Please help me, it's really annoying question i'm practicing and it sucks. Thanks everyone.

Answer 1

Note: Keep in mind that velocity takes into account direction so going up 20.4 m/s is positive and going down 20.4 m/s is -20.4 m/s

To solve the first part you are

Given: V0= 39.8 m/s, Vf= 20.4 m/s, g= -9.81 m.s^2,

so vf=v0+gt

20.4 m/s = 39.8 m/s+(-9.81 m.s^2)t

-19.4 m/s= (- 9.81 m/s^2)t

1.98 s=t

For the second part

Given:v0=39.8 m/s, vf=-20.4 m/s, g= -9.81 m.s^2,

so vf=v0+gt

-20.4 m/s=39.8 m/s+(-9.81 m.s^2)t

-60.2 m/s=(-9.81 m.s^2)t

6.14 s=t

From here you should be able to solve the rest

Answer 2

Remember that acceleration means the rate at which velocity changes.  When something accelerates due to gravity (near the earth's surface), its vertical speed changes by 9.8 m/s every second.  While it's rising, its speed DECREASES by 9.8 m/s every second; and while it's falling its speed INCREASES by 9.8 m/s every second.

So: it starts at 39.8 m/s.

After 1 second, it has slowed to 39.8 - 9.8 = 30.0 m/s (still going up)

After 2 seconds, it has slowed to 30.0 - 9.8 = 20.2 m/s (still going up)

So, the answer to the first question is: "about 2 seconds."  to get the exact answer:

t = (change in velocity) / acceleration

= (final velocity - initial velocity) / acceleration

= (20.4 m/s - 39.8 m/s) / (-9.8 m/s²)

= 1.98 seconds

(I used a negative sign for the acceleration (-9.8 m/s²) to indicate that it points DOWNWARD.  I'm using the convention that up=positive and down=negative.)

> At what time is it moving at a speed 20.4 m/s DOWNWARD?

Use the exact same formula, only this time the "final velocity" is "-24 m/s" (negative because it's heading downward):

t = (-20.4 m/s - 39.8 m/s) / (-9.8 m/s²)

> When is the displacement of the boulder from its initial position zero?

For this you need to know that when it reaches ground level, it will be going at the same speed as when it initially shot up.  So use the same formula, but now "final velocity" is "-39.8 m/s" (negative because it's heading downward):

t = (-39.8 m/s - 39.8 m/s) / (-9.8 m/s²)

> When is the velocity of the boulder zero?

Exact same formula, but this time the "final velocity" is zero:

t = (0 m/s - 39.8 m/s) / (-9.8 m/s²)

> What is the magnitude of the acceleration while the boulder is moving?

For ANYTHING that's moving only under the influence of gravity (near the surface of the earth), the acceleration's magnitute is always 9.8 m/s²

> What is the direction of the acceleration while the boulder is moving?

The acceleration of gravity is in the same direction as the force of gravity: downward.