Thread: rate

**John 5** · July 5th 2007, 02:31 PM

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?

**Jonboy** · July 5th 2007, 03:17 PM

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

**Soroban** · July 5th 2007, 03:27 PM

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/sec² or 4.9 m/sec².

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.

**rualin** · July 5th 2007, 03:31 PM

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...

**Jonboy** · July 5th 2007, 03:41 PM

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.

**topsquark** · July 5th 2007, 03:58 PM

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

**rualin** · July 5th 2007, 04:47 PM

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

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