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Okay, so I need to solve this problem where a car accelerates from rest with a constant acceleration, and after 10 seconds, its velocity is 20 m/s. Then, it decelerates at a constant rate of 2 m/s² until it comes to a stop. The question is asking for the total distance traveled. Hmm, let me break this down step by step.
First, the car starts from rest, so initial velocity u1 is 0 m/s. It accelerates with constant acceleration a1 for 10 seconds, reaching a velocity v1 of 20 m/s. Then, it decelerates at a constant rate a2 of -2 m/s² until it stops. I need to find the total distance, which would be the sum of the distance covered during acceleration and the distance covered during deceleration.
Let me start with the first part: acceleration phase. I remember the formula for velocity under constant acceleration is v = u + at. Here, u1 is 0, v1 is 20 m/s, and t is 10 seconds. So plugging in the values: 20 = 0 + a1*10. Solving for a1 gives a1 = 20 / 10 = 2 m/s². So the acceleration is 2 m/s².
Now, to find the distance during acceleration, I can use the formula s = ut + 0.5at². Using the values for the first phase: s1 = 010 + 0.52*(10)². That simplifies to 0 + 0.52100 = 0.5*200 = 100 meters. So the car travels 100 meters during the acceleration phase.
Next, the deceleration phase. The car starts at velocity v1 = 20 m/s and decelerates at a2 = -2 m/s² until it stops, so final velocity v2 = 0 m/s. I need to find the time it takes to stop and the distance covered during this time.
Using the same velocity formula: v = u + at. Here, u2 = 20 m/s, v2 = 0, a2 = -2 m/s². So 0 = 20 + (-2)*t. Solving for t: -20 = -2t => t = 10 seconds. So it takes 10 seconds to decelerate to a stop.
Now, the distance during deceleration can be found using s = ut + 0.5at². Plugging in the values: s2 = 2010 + 0.5(-2)(10)². Calculating that: 200 + 0.5(-2)*100 = 200 - 100 = 100 meters. Wait, so the deceleration phase also results in 100 meters?
Alternatively, maybe I can use another formula that doesn't require time, like v² = u² + 2as. Let me check that. For the deceleration phase: final velocity squared is 0, initial velocity squared is (20)^2 = 400. So 0 = 400 + 2*(-2)*s2. Solving for s2: 0 = 400 -4s2 => 4s2 = 400 => s2 = 100 meters. Yep, same result. So that's consistent.
Therefore, total distance is s1 + s2 = 100 + 100 = 200 meters.
Wait, but let me double-check everything. The acceleration phase: 10 seconds at 2 m/s² starting from 0, reaching 20 m/s. The distance is 100 meters. Then deceleration phase: 20 m/s to 0 with -2 m/s², taking 10 seconds and 100 meters. Total 200 meters. That seems right.
Alternatively, maybe I can visualize the velocity-time graph. The area under the graph gives the distance. For the first 10 seconds, it's a triangle with base 10 and height 20. Area is 0.51020 = 100. Then the next 10 seconds, it's a triangle with base 10 and height 20 again, so another 100. Total area 200. Yep, same answer.
So I think the total distance traveled is 200 meters.
The total distance traveled by the car is calculated by summing the distances during acceleration and deceleration phases.
Acceleration Phase:
Initial velocity, ( u_1 = 0 , \text{m/s} )
Final velocity, ( v_1 = 20 , \text{m/s} )
Time, ( t_1 = 10 , \text{s} )
Acceleration, ( a_1 = \frac{v_1 - u_1}{t_1} = \frac{20}{10} = 2 , \text{m/s}^2 )
Distance: ( s_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2 = 0 + \frac{1}{2} \cdot 2 \cdot 10^2 = 100 , \text{m} )
Deceleration Phase:
Initial velocity, ( u_2 = 20 , \text{m/s} )
Final velocity, ( v_2 = 0 , \text{m/s} )
Deceleration, ( a_2 = -2 , \text{m/s}^2 )
Time to stop: ( t_2 = \frac{v_2 - u_2}{a_2} = \frac{-20}{-2} = 10 , \text{s} )
Distance: ( s_2 = u_2 t_2 + \frac{1}{2} a_2 t_2^2 = 20 \cdot 10 + \frac{1}{2} \cdot (-2) \cdot 10^2 = 200 - 100 = 100 , \text{m} )
Total Distance:
[
s_{\text{total}} = s_1 + s_2 = 100 , \text{m} + 100 , \text{m} = \boxed{200 , \text{meters}}
]
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发表于 2025-11-6 17:20:00