1. ## rate

1. (a) A car traveling at a speed of v can brake to an emergency stop in a distance x. Assuming all other driving conditions arte all similar, if traveling speed of the car doubles, the stopping distance will be (1) 2x, or (2) 2x, or (3) 4x: (b) a driver traveling at 40.0 km/h in a school zone can brake to an emergency stop in 3.00 m. What would be the braking distance if the car were traveling at 60.0 km/h?

2. A student drops a ball from the top of a building; it takes 2.8 s for the ball to reach the ground. (a) What is the ball's speed just before hitting the ground? (b) What is the height of the building?

3. A photographer in a helicopter ascending vertically at a constant rate of 12.5 m/s accidentally drops a camera out the window when the helicopter is 60 m above the ground. (a) How long will it take the camera to reach the ground? (b) What will it's speed be when it hits?

2. I'm sure this requires nothing more than some formulas.

3. Hello, John!

For the last two, we are expected to know the "free fall" formula: . $y \; = \; h_o + v_ot - gt^2$

. . where: . $\begin{array}{ccc}h_o & = & \text{initial height} \\ v_o & = &\text{initial velocity} \\ g & = & \text{gravitational constant} \\ y & = & \text{height of object} \end{array}$

Note: $g$ is usually 16 ft/secē or 4.9 m/secē.

2. A student drops a ball from the top of a building;
it takes 2.8 s for the ball to reach the ground.
(a) What is the ball's speed just before hitting the ground?
(b) What is the height of the building?
We have: . $v_o = 0$ . (the ball is dropped, not thrown).

The equation is: . $y \;=\;h_o - 16t^2$

(a) The velocity is given by the derivative: . $v(t) \:=\:y' \:=\:-32t$

Then: . $v(2.8) \:=\:-32(2.8) \:=\:-89.6$

The ball's speed on impact is 89.6 feet per second (downward, of course).

(b) When the ball strikes the ground, its height is zero.
. . We have: . $0 \;=\;h_o - 16\cdot2.8^2\quad\Rightarrow\quad h_o \:=\:125.44$

The height of the building is 125.44 feet.

3. A photographer in a helicopter ascending vertically at a constant rate of 12.5 m/s
accidentally drops a camera out the window when the helicopter is 60 m above the ground.
(a) How long will it take the camera to reach the ground?
(b) What will its speed be when it hits?
We are given: . $h_o = 60,\;v_o = 12.5$

The equation is: . $y \;=\;60 + 12.5t - 4.9t^2$

(a) "reach the ground" means: $y = 0$
We have: . $60 + 12.5t - 4.9t^2 \:=\:0$
. . which factors: . $(t - 5)(4.9t + 12) \:=\:0$
. . and has the positive root: . $t = 5$
Therefore, it takes the camera 5 seconds to crash to the ground.

(b) The velocity is: . $v(t) \;=\;y' \;=\;12.5 - 9.8t$
. . Hence: . $v(5)\;=\;12.5 - 9.8(5) \;=\;-36.5$
Therefore, its speed on impact is 36.5 meters per second.

4. aaaand...

Problem 1,a
We are told that the square of velocity is proportional to the stopping distance. (look at topsquark's post for its justification)
$v^2 \propto x$ so $(2v)^2=4v \propto 4x$.

4x

Problem 1,b
$40^2=1600\frac{km}{h} \propto 3.00m$

Using this information, we can find the stopping distance of the same car at 60.0 km/h by using a proportion and solving for x.

$\frac{3}{1600} = \frac{x}{3600} \Longleftrightarrow x=6.75m$

These veterans are just too fast...

5. Originally Posted by rualin
These veterans are just too fast...
Soroban is not the fastest so you haven't seen anything yet lol j/m. Soroban is one of the most cunning and interesting.

6. Originally Posted by rualin
Problem 1,a
We are told that velocity is directly proportional to the stopping distance.
No we weren't. Try this programme:
Find the acceleration for when it starts at speed v.
The acceleration will be the same as this for when it starts at 2v.

In both cases, apply the formula:
$v^2 = v_0^2 + 2a(x - x_0)$

-Dan

7. Thanks, topsquark! I realize my mistake and have corrected it... I hope.

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### A photographer in a helicopter ascending vertically at a constant rate of 12.5 m/s accidentally drops a camera out the window when the helicopter is 60.0 m above the ground.

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