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Okay, so I need to figure out how to solve this problem where a car accelerates from rest with a constant acceleration, and after 10 seconds, it reaches a speed of 20 m/s. Then, I have to find the distance it travels during the first 5 seconds. Hmm, let's start by recalling some basic kinematics equations.
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First, the car starts from rest, so the initial velocity, u, is 0 m/s. The acceleration is constant, which is good because that means I can use the standard equations. After 10 seconds, the velocity is 20 m/s. I think the first step is to find the acceleration. The equation that relates velocity, acceleration, and time is v = u + at. Since u is 0, that simplifies to v = at. Plugging in the values I have: 20 m/s = a * 10 s. Solving for a, I divide both sides by 10, so a = 20 / 10 = 2 m/s². Okay, so the acceleration is 2 meters per second squared.
Now, I need to find the distance traveled during the first 5 seconds. For this, I can use another kinematic equation. Since we're dealing with constant acceleration and we know the initial velocity, time, and acceleration, the equation that comes to mind is s = ut + 0.5 * a * t². Again, u is 0, so the first term drops out, and we have s = 0.5 * a * t². Plugging in the values: s = 0.5 * 2 m/s² * (5 s)². Let me compute that step by step.
First, 0.5 multiplied by 2 is 1. Then, 5 squared is 25. So, 1 * 25 = 25. Therefore, the distance traveled in the first 5 seconds is 25 meters. Wait, let me double-check that. If the acceleration is 2 m/s², then after 5 seconds, the velocity would be v = at = 2*5 = 10 m/s. The average velocity during those 5 seconds would be (initial + final)/2 = (0 + 10)/2 = 5 m/s. Then, distance is average velocity multiplied by time, so 5 m/s * 5 s = 25 meters. Yep, that matches. So both methods give the same result, which is reassuring.
Alternatively, I could use another equation: s = (v² - u²)/(2a). Since u is 0, that simplifies to s = v²/(2a). But wait, here v would be the final velocity at 5 seconds, which we found to be 10 m/s. Plugging in, s = (10)^2 / (2*2) = 100 / 4 = 25 meters. Again, same answer. So all three equations confirm that the distance is 25 meters.
Just to make sure I didn't mix up any numbers, let's recap. Acceleration found from the first part: 2 m/s². Then using that acceleration to compute distance in 5 seconds. All steps check out. I think that's solid. No mistakes noticed here. The key was remembering the right kinematic equations and plugging in the values correctly. It's easy to mix up equations, but cross-verifying with different methods helps ensure accuracy.
To solve the problem, we first determine the car's acceleration using the given information and then calculate the distance traveled in the first 5 seconds.
Step 1: Find the acceleration (a)
Initial velocity, ( u = 0 , \text{m/s} )
Final velocity, ( v = 20 , \text{m/s} )
Time, ( t = 10 , \text{s} )
Using the equation ( v = u + at ):
[
20 = 0 + a \cdot 10 \implies a = \frac{20}{10} = 2 , \text{m/s}^2
]
Step 2: Calculate the distance traveled in the first 5 seconds (s)
Using the equation ( s = ut + \frac{1}{2}at^2 ) with ( u = 0 ):
[
s = 0 \cdot 5 + \frac{1}{2} \cdot 2 \cdot (5)^2 = 0 + 1 \cdot 25 = 25 , \text{m}
]
Verification using alternative methods:
Average velocity method:
Final velocity at 5 s: ( v = at = 2 \cdot 5 = 10 , \text{m/s} )
Average velocity: ( \frac{0 + 10}{2} = 5 , \text{m/s} )
Distance: ( 5 , \text{m/s} \cdot 5 , \text{s} = 25 , \text{m} )
Kinematic equation ( s = \frac{v^2 - u^2}{2a} ):
( s = \frac{(10)^2 - 0}{2 \cdot 2} = \frac{100}{4} = 25 , \text{m} )
Final Answer: The car travels (\boxed{25}) meters in the first 5 seconds.
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