As soon as a traffic light turns green, a car speeds up from rest to 50.0 mi/h with a constant acceleration of 9.90 mi/h·s. In the adjoining bike lane, a cyclist speeds up from rest to 20.0 mi/h with a constant acceleration of 13.5 mi/h·s. Each vehicle maintains a constant velocity after reaching its cruising speed.
(a) For how long is the bicycle ahead of the car?
(b) By what maximum distance does the bicycle lead the car?
I ended up with a quadratic equation but the answer was incorrect, but it was so long to solve that I don't know at what point I would go wrong.
Solving a quadratic equation can indeed be time-consuming and error-prone, especially if you made a mistake during the calculation. However, there is a more efficient way to solve this problem by using kinematic equations. Let's go through the steps to find the correct answers.
(a) To determine for how long the bicycle is ahead of the car, we need to find the time it takes for each vehicle to reach their cruising speeds.
Let's start with the car. We can use the equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity (0 mph in this case as the car is at rest), a is the acceleration (9.90 mi/h·s), and t is the time.
Plugging in the values, we have:
50.0 mi/h = 0 mi/h + (9.90 mi/h·s) * t
Rearranging the equation, we get:
t = 50.0 mi/h / (9.90 mi/h·s)
t ≈ 5.05 seconds
Therefore, it takes the car approximately 5.05 seconds to reach 50.0 mi/h.
Now let's calculate the time for the bicycle:
20.0 mi/h = 0 mi/h + (13.5 mi/h·s) * t
Rearranging the equation, we get:
t = 20.0 mi/h / (13.5 mi/h·s)
t ≈ 1.48 seconds
Therefore, it takes the bicycle approximately 1.48 seconds to reach 20.0 mi/h.
Since the bicycle reaches its cruising speed faster than the car, we can conclude that the bicycle is ahead of the car for a shorter amount of time. Therefore, the bicycle is ahead of the car for approximately 1.48 seconds.
(b) To find the maximum distance by which the bicycle leads the car, we can use the equation:
d = vi*t + 0.5*a*t^2
where d is the distance, vi is the initial velocity (0 mph in this case as both vehicles start from rest), a is the acceleration, and t is the time.
For the car:
d_car = 0.5 * (9.90 mi/h·s) * (5.05 seconds)^2
d_car ≈ 125.25 miles
For the bicycle:
d_bike = 0.5 * (13.5 mi/h·s) * (1.48 seconds)^2
d_bike ≈ 14.49 miles
Therefore, the bicycle leads the car by a maximum distance of approximately 14.49 - 125.25 = -110.76 miles.
It's important to note that the negative sign indicates that the car is ahead of the bicycle by 110.76 miles, rather than the other way round. This means that the car overtakes and passes the bicycle during the motion.
By following these steps and avoiding errors in calculations, you can obtain the correct answers efficiently without having to solve a quadratic equation directly.
To solve this problem, let's break it down step by step:
Step 1: Convert all given quantities to SI units (meters and seconds) to make the calculations easier.
Given:
Initial velocity of the car (u₁) = 0 mi/h
Final velocity of the car (v₁) = 50.0 mi/h
Acceleration of the car (a₁) = 9.90 mi/h·s
Given:
Initial velocity of the bicycle (u₂) = 0 mi/h
Final velocity of the bicycle (v₂) = 20.0 mi/h
Acceleration of the bicycle (a₂) = 13.5 mi/h·s
We need to convert these velocities and accelerations to SI units:
1 mi/h = 0.44704 m/s
1 mi/h·s = 0.44704 m/s²
So, let's convert the given quantities:
v₁ = 50.0 mi/h × 0.44704 m/s = 22.352 m/s
a₁ = 9.90 mi/h·s × 0.44704 m/s² = 4.423 m/s²
v₂ = 20.0 mi/h × 0.44704 m/s = 8.9408 m/s
a₂ = 13.5 mi/h·s × 0.44704 m/s² = 6.04224 m/s²
Step 2: Calculate the time it takes for each vehicle to reach its cruising speed.
Use the formula:
v = u + at
For the car:
u₁ = 0 m/s (starting from rest)
v₁ = 22.352 m/s (cruising speed)
a₁ = 4.423 m/s²
Rearrange the formula to solve for time (t):
t₁ = (v₁ - u₁) / a₁
t₁ = (22.352 m/s - 0 m/s) / 4.423 m/s²
t₁ ≈ 5.05 s
For the bicycle:
u₂ = 0 m/s (starting from rest)
v₂ = 8.9408 m/s (cruising speed)
a₂ = 6.04224 m/s²
Rearrange the formula to solve for time (t):
t₂ = (v₂ - u₂) / a₂
t₂ = (8.9408 m/s - 0 m/s) / 6.04224 m/s²
t₂ ≈ 1.48 s
Step 3: Calculate the distance traveled by each vehicle during the time it takes to reach its cruising speed.
Use the formula:
s = ut + 0.5at²
For the car:
u₁ = 0 m/s
t₁ = 5.05 s
a₁ = 4.423 m/s²
s₁ = u₁t₁ + 0.5a₁t₁²
s₁ = 0 m/s × 5.05 s + 0.5 × 4.423 m/s² × (5.05 s)²
s₁ ≈ 55.96 m
For the bicycle:
u₂ = 0 m/s
t₂ = 1.48 s
a₂ = 6.04224 m/s²
s₂ = u₂t₂ + 0.5a₂t₂²
s₂ = 0 m/s × 1.48 s + 0.5 × 6.04224 m/s² × (1.48 s)²
s₂ ≈ 5.34 m
Step 4: Answer the questions:
(a) For how long is the bicycle ahead of the car?
The bicycle is ahead of the car for the time it takes for the car to reach its cruising speed, which is t₁ ≈ 5.05 s.
(b) By what maximum distance does the bicycle lead the car?
The maximum distance the bicycle leads the car is the difference between the distances traveled by each vehicle during the time it takes for the car to reach its cruising speed.
Difference = s₁ - s₂
Difference ≈ 55.96 m - 5.34 m
Difference ≈ 50.62 m
Therefore, the bicycle leads the car by a maximum distance of approximately 50.62 meters.