How far up the hill will it coast before starting to roll back down A car traveling at 23.0m/s runs out of gas while traveling up a 25.0 slope

How far up the hill will it coast before starting to roll back down A car traveling at 23.0m/s runs out of gas while traveling up a 25.0 slope

The correct answer and explanation is:

Let’s solve the problem step-by-step.

Problem:

A car traveling at 23.0 m/s runs out of gas while going up a hill inclined at 25.0°. How far up the hill will the car coast before it starts rolling back down?

What is being asked?

We need to find the distance along the slope (the hill) that the car will coast upward before it stops and starts rolling back.

Key information:

• Initial speed, v0=23.0 m/sv_0 = 23.0 \, m/s
• Angle of slope, θ=25.0∘\theta = 25.0^\circ
• Final speed at the highest point, v=0 m/sv = 0 \, m/s (the car stops momentarily before rolling back)
• Gravity acceleration, g=9.8 m/s2g = 9.8 \, m/s^2
• We ignore friction and air resistance for this calculation.

Step 1: Analyze the forces and acceleration

When the car is coasting up the hill, it slows down due to gravity pulling it back downhill.

The component of gravitational acceleration along the slope is: a=gsin⁡θa = g \sin \theta

Since the car is moving upward but slowing down, acceleration is negative relative to velocity direction: a=−gsin⁡θa = -g \sin \theta

Calculate aa: a=−9.8×sin⁡(25∘)≈−9.8×0.4226=−4.14 m/s2a = -9.8 \times \sin(25^\circ) \approx -9.8 \times 0.4226 = -4.14 \, m/s^2

Step 2: Use kinematic equation

We use the kinematic equation to find the distance dd the car travels up the slope until it stops: v2=v02+2adv^2 = v_0^2 + 2 a d

Since final velocity v=0v = 0, rearranged: 0=v02+2ad  ⟹  d=−v022a0 = v_0^2 + 2 a d \implies d = – \frac{v_0^2}{2 a}

Note the negative sign of acceleration cancels out with negative sign in numerator, so distance is positive.

Calculate dd: d=−(23.0)22×(−4.14)=5298.28≈63.9 md = – \frac{(23.0)^2}{2 \times (-4.14)} = \frac{529}{8.28} \approx 63.9 \, m

Final answer:

The car will coast approximately 63.9 meters up the hill before it stops and starts rolling back down.

Explanation (300 words):

When a car runs out of gas while moving uphill, it no longer has the engine force to counteract gravity, so it will coast upward, gradually losing speed until it momentarily stops. The distance it covers before rolling back is determined by its initial speed and the hill’s slope.

The key physics concept here is the component of gravitational acceleration along the slope. Gravity acts vertically downward at 9.8 m/s², but on an inclined plane, only a part of this force works against the motion. This component equals gsin⁡θg \sin \theta, where θ\theta is the angle of the hill. For a 25° slope, this component slows the car by about 4.14 m/s².

To find how far the car will coast, we use kinematic equations for motion with constant acceleration. The car’s initial speed is 23 m/s uphill, and it slows down uniformly at 4.14 m/s² until the speed becomes zero. Using the formula v2=v02+2adv^2 = v_0^2 + 2 a d, where final velocity v=0v = 0, we solve for dd, the distance traveled.

Plugging in the values gives a distance of about 64 meters. This distance represents how far up the hill the car will coast before gravity overcomes the initial kinetic energy, stopping the car momentarily. After this, the car will start rolling back downhill under gravity.

This calculation assumes no friction or air resistance, which in real life would reduce the distance somewhat. Nevertheless, it provides a good estimate of the car’s maximum coasting distance uphill after running out of fuel.