A car is traveling along a straight road at a velocity of +36.0 m/s when its engine cuts out. For the next ten seconds, the car slows down, and its average acceleration is 1. For the next five seconds, the car slows down further, and its average acceleration is 2. The velocity of the car at the end of the fifteen-second period is +21.5 m/s. The ratio of the average acceleration values is 1/2 = 1.75. Find the velocity of the car at the end of the initial ten-second interval
If the car is slowing down, the acceleration would be negative. Your statement 1/2 = 1.75 also does not make sense. Please make sure you are stating the problem correctly.
Well, it seems like I mixed up some numbers there. My sincerest apologies for the confusion. Let's try to straighten things out, shall we?
Given the information provided, we know that the car's initial velocity is +36.0 m/s and that its average acceleration during the first ten seconds is 1 m/s². We are asked to find the velocity of the car at the end of this ten-second interval.
To find the final velocity, we can use the equation:
v = u + at,
where:
v is the final velocity,
u is the initial velocity,
a is the average acceleration, and
t is the time.
In this case, u = 36.0 m/s, a = 1 m/s², and t = 10 s.
Plugging these values into the equation, we get:
v = 36.0 m/s + (1 m/s²) * 10 s = 36.0 m/s + 10 m/s = 46.0 m/s.
Therefore, the velocity of the car at the end of the initial ten-second interval is 46.0 m/s.
Hope that clears the air! Let me know if you have any more questions.
Apologies for the confusion caused earlier. Let's solve the problem step by step using the corrected information:
Given:
Initial velocity (v₁) = +36.0 m/s
Final velocity (v₂) = +21.5 m/s
Acceleration in the first 10 seconds (a₁) = -1 m/s²
Acceleration in the next 5 seconds (a₂) = -2 m/s²
To find: Velocity at the end of the initial ten-second interval (vₓ)
We can use the equation:
v₂ = v₁ + a₁ * t₁ + a₂ * t₂
Where:
t₁ = 10 seconds (duration of the first 10 seconds)
t₂ = 5 seconds (duration of the next 5 seconds)
Substituting the given values, we have:
+21.5 m/s = +36.0 m/s + (-1 m/s²) * 10s + (-2 m/s²) * 5s
Now, let's compute:
vₓ = 21.5 m/s - 36.0 m/s + (1 m/s²) * 10s + (2 m/s²) * 5s
vₓ = -14.5 m/s + 10 m/s + 10 m/s
vₓ = -14.5 m/s + 20 m/s
vₓ = +5.5 m/s
Therefore, the velocity of the car at the end of the initial ten-second interval is +5.5 m/s.
I apologize for the confusion. Let's restate the problem correctly and solve it step by step.
Given:
- Initial velocity of the car, v₀ = +36.0 m/s
- Average acceleration for the first 10 seconds, a₁ = -1 m/s² (negative because the car is slowing down)
- Average acceleration for the next 5 seconds, a₂ = -2 m/s²
- Final velocity at the end of the 15-second period, v₃ = +21.5 m/s
To find the velocity of the car at the end of the initial ten-second interval, we need to determine the velocity at the 10-second mark. Let's break down the problem into smaller steps:
Step 1: Find the change in velocity during the first 10 seconds.
Using the average acceleration and time for the first 10 seconds, we can use the equation:
Δv₁ = a₁ * t₁
Where Δv₁ is the change in velocity during the first 10 seconds, a₁ is -1 m/s², and t₁ is 10 seconds.
Δv₁ = -1 m/s² * 10 s
Δv₁ = -10 m/s
So, the velocity change during the first 10 seconds is -10 m/s.
Step 2: Find the velocity at the 10-second mark.
To find the velocity at the 10-second mark, we need to subtract the velocity change from the initial velocity.
v₁₀ = v₀ + Δv₁
Where v₁₀ is the velocity at the 10-second mark, v₀ is +36.0 m/s, and Δv₁ is -10 m/s.
v₁₀ = +36.0 m/s - 10 m/s
v₁₀ = +26.0 m/s
So, the velocity of the car at the end of the initial ten-second interval is +26.0 m/s.