Question

A truck travels at a constant speed of 28 m/s. It arrives at the top of a 65 meter hill and then COASTS (taking the foot off the gas) down the hill to a valley that is 50 meters lower than the hill. What is the car’s speed in the valley?

Answer 1

To determine the truck’s speed in the valley when it coasts down a hill, we can use energy conservation to show that the final speed depends on both the initial speed at the top of the hill and the height of the descent.

Step-by-step explanation:

You are asking about the speed of a truck as it descends from a hill to a valley. When the truck coasts down the hill, it is subject to gravitational acceleration, which increases its speed due to the conversion of potential energy into kinetic energy. To find the car’s speed in the valley, we can use the principle of conservation of energy.

Assuming no friction and air resistance, the initial potential energy at the top of the hill plus the initial kinetic energy will equal the kinetic energy at the bottom.

Let’s denote the mass of the truck as m, gravitational acceleration as g (9.8 m/s²), the height of the hill as h, and the initial and final velocities as vᵢ and vᵣ respectively. The energy conservation equation is given by:

mgh + ½ mvᵢ² = ½ mvᵣ²

Since the truck is coasting, this simplifies to:

vᵣ = √(vᵢ² + 2gh)

Plugging in the given values:

vᵣ = √((28 m/s)² + 2 * 9.8 m/s² * 50 m)

The final calculation will give the truck's speed in the valley.

Answer 2

Approximately
`35.7\; {\rm m\cdot s^(-1)}` (Assumptions:
`g = 9.81\; {\rm m\cdot s^(-2)}`; the effect of friction is negligible; mass of the wheels is negligible.)

Step-by-step explanation:

Under the assumptions, while the vehicle is coasting down the hill, gravitational potential energy
`(\text{GPE})` of the vehicles is converted into kinetic energy
`(\text{KE})`. Since the vehicle was moving at a certain velocity at the top of hill, the energy converted from
`(\text{GPE})\!` would be added to the initial value of
`(\text{KE})\!`:

`(\text{KE, current}) = (\text{KE, initial}) + (\text{GPE})`.

Let
`m` denote the mass of the truck. Let
`v_(0)` denote the velocity of the vehicle at the top of the hill, and let
`v_(1)` denote the velocity of the vehicle in the valley. Let
`\Delta h` denote the height difference between the valley and the top of the hill.

While the vehicle is at the top of the hill, velocity of the vehicle was
`v_(0) = 28\; {\rm m\cdot s^(-1)}`. The kinetic energy of this vehicle at that instant would be:

`\displaystyle (\text{KE, initial}) = (1)/(2)\, m\, {v_(0)}^(2)`.

The height difference between the valley and the top of the hill is
`\Delta h = 65\; {\rm m}`. During this motion, the change in the
`(\text{GPE})` of the vehicle would be:

`(\text{GPE}) = m\, g\, \Delta h`.

Hence, when the vehicle reached the valley, kinetic energy of the vehicle would be:

`\begin{aligned} (\text{KE, current}) &= (\text{KE, initial}) + (\text{GPE}) \\ &= (1)/(2)\, m\, {v_(0)}^(2) + m\,g \, \Delta h\end{aligned}`.

Since
`(\text{KE, current}) = (1/2)\, m\, {v_(1)}^(2)`, the equation above would be equivalent to:

`\begin{aligned} (1)/(2)\, m\, {v_(1)}^(2)= (1)/(2)\, m\, {v_(0)}^(2) + m\,g \, \Delta h\end{aligned}`.

Rearrange this equation to find
`v_(1)`, the velocity of the vehicle when it reaches the valley:

`\begin{aligned}v_(1) &= \sqrt{\frac{\displaystyle 2\, \left((1/2)\, m\, {v_(0)}^(2) + m\, g\, \Delta h\right)}{m}} \\ &= \sqrt{\displaystyle 2\, \left((1/2){v_(0)}^(2) +g\, \Delta h\right)} \\ &= \sqrt{\displaystyle {v_(0)}^(2) + 2\, g\, \Delta h} \\ &= \sqrt{\displaystyle (28\;{\rm m\cdot s^(-1)})^(2) + 2\, (9.81\; {\rm m\cdot s^(-2)})\, (65\; {\rm m})} \\ &\approx 42.0\; {\rm m\cdot s^(-1)}\end{aligned}`.

In other words, the velocity of the vehicle would be approximately
`42.0\; {\rm m\cdot s^(-1)}` when it reaches the valley.