PRE

CasinoGames

Okay, so I need to solve this problem where a car accelerates from rest with a constant acceleration of 2 m/s². The question is asking for the distance traveled in the first 5 seconds. Hmm, let's see. I remember there are some kinematic equations for motion under constant acceleration. Let me recall them.


First, the basic equations. When an object starts from rest, its initial velocity is zero. The acceleration is given as 2 m/s². Time is 5 seconds. We need to find the distance. The formula that comes to mind is the one that relates distance, initial velocity, acceleration, and time. I think it's s = ut + (1/2)at². Let me check if that's right.


Yes, s is the displacement, u is the initial velocity, a is acceleration, and t is time. Since the car starts from rest, u = 0. So the equation simplifies to s = (1/2)at². Plugging in the values, a = 2 m/s² and t = 5 seconds. Let me compute that.


First, square the time: 5 squared is 25. Multiply that by acceleration: 25 * 2 = 50. Then multiply by 1/2: 50 * 0.5 = 25. So the distance should be 25 meters. Wait, that seems straightforward, but let me make sure I didn't mix up any formulas.


Alternatively, I can use another kinematic equation to verify. Maybe the one involving final velocity. The final velocity v = u + at. Since u is 0, v = at. So after 5 seconds, v = 2*5 = 10 m/s. Then, average velocity would be (initial + final)/2 = (0 + 10)/2 = 5 m/s. Distance is average velocity times time: 5 m/s * 5 s = 25 meters. Same result. Okay, that checks out.




Wait, another way to think about it: since the acceleration is constant, the distance-time graph would be a parabola. The area under the velocity-time graph gives the distance. The velocity starts at 0 and increases linearly to 10 m/s at t=5. The area under that graph is a triangle with base 5 and height 10. Area = (1/2)baseheight = (1/2)510 = 25 meters. Yep, same answer.


So all three methods give me 25 meters. That seems consistent. I don't think I made a mistake here. Let me just recap the steps to be thorough.


Identify known quantities: initial velocity (u) = 0, acceleration (a) = 2 m/s², time (t) = 5 s.
Choose the appropriate kinematic equation. Since we need distance and have time, acceleration, and initial velocity, the equation s = ut + (1/2)at² is suitable.
Substitute the values into the equation: s = 05 + 0.52*(5)².
Calculate step by step: 5 squared is 25, multiplied by 2 gives 50, multiplied by 0.5 gives 25.
Verified using alternative methods (final velocity and average velocity, and area under velocity-time graph) to ensure correctness.


All steps lead to the same conclusion. Therefore, the distance traveled by the car in the first 5 seconds is 25 meters.




The car starts from rest (( u = 0 )) with a constant acceleration (( a = 2 , \text{m/s}^2 )) and travels for ( t = 5 , \text{s} ). To find the distance (( s )):


Use the kinematic equation:
[
s = ut + \frac{1}{2}at^2
]
Substitute the known values:
[
s = 0 \cdot 5 + \frac{1}{2} \cdot 2 \cdot (5)^2
]
[
s = 0 + \frac{1}{2} \cdot 2 \cdot 25
]
[
s = 25 , \text{meters}
]


Verification using average velocity:


Final velocity: ( v = u + at = 0 + 2 \cdot 5 = 10 , \text{m/s} )
Average velocity: ( \frac{u + v}{2} = \frac{0 + 10}{2} = 5 , \text{m/s} )
Distance: ( \text{Average velocity} \times \text{Time} = 5 , \text{m/s} \times 5 , \text{s} = 25 , \text{m} )


Verification using area under velocity-time graph:


The graph forms a triangle with base ( 5 , \text{s} ) and height ( 10 , \text{m/s} ):
[
\text{Area} = \frac{1}{2} \times 5 \times 10 = 25 , \text{m}
]


Final Answer:

The distance traveled by the car in the first 5 seconds is (\boxed{25}) meters.