this is name 77 2025-03-25 Algebra 0 views Problem Diagram Problem Statement this is des 77 Solution Process Okay, let's see. The student uploaded a math problem. Let me take a closer look. The problem says: "A car is moving at a constant speed of 20 meters per second. It decelerates uniformly for 5 seconds until it comes to rest. Determine the acceleration of the car and the distance it travels during the deceleration." Alright, first, I need to remember the basic equations of motion for constant acceleration. The key here is that the deceleration is uniform, which means the acceleration is constant. Since the car comes to rest, the final velocity will be 0 m/s. The initial velocity is given as 20 m/s, and the time taken to stop is 5 seconds. So, to find acceleration, I can use the equation: acceleration = (final velocity - initial velocity) / time. Plugging in the numbers: (0 - 20)/5 = -4 m/s². The negative sign indicates deceleration. That seems straightforward. Now for the distance traveled during deceleration. There are a couple of equations I can use here. One is distance = average velocity × time. Average velocity when acceleration is constant is (initial + final)/2. So, (20 + 0)/2 = 10 m/s. Multiply that by 5 seconds gives 50 meters. Alternatively, using the equation s = ut + ½at², where u is initial velocity, a is acceleration, and t is time. Plugging in: 20*5 + 0.5*(-4)*(5)² = 100 - 50 = 50 meters. Both methods give the same result, which is good. Wait, maybe I should check if there's another way to confirm the distance. Using the equation v² = u² + 2as, since final velocity v is 0, we can solve for s: 0 = 20² + 2*(-4)*s → 400 = 8s → s = 50 meters. Yep, same answer. So all three methods confirm the distance is 50 meters. That's solid. I should make sure the student understands why the acceleration is negative. It's because the car is slowing down, so the acceleration vector is opposite to the direction of motion. Also, emphasizing units is important—acceleration in m/s² and distance in meters. Are there any common mistakes the student might make here? Maybe mixing up initial and final velocities, or forgetting the negative sign for deceleration. Also, when using the equations, ensuring they use the correct signs consistently. For example, if they take acceleration as positive, they might get confused. So reinforcing the sign convention is key. **Step-by-Step Explanation:** 1. **Identify Known Values:** - Initial velocity (\(u\)) = 20 m/s - Final velocity (\(v\)) = 0 m/s (car comes to rest) - Time (\(t\)) = 5 seconds 2. **Find Acceleration (\(a\)):** Use the formula: \[ a = \frac{v - u}{t} \] Plug in the values: \[ a = \frac{0 - 20}{5} = -4 \, \text{m/s}^2 \] The negative sign means the car is decelerating. 3. **Find Distance Traveled (\(s\)):** Use the formula for distance with constant acceleration: \[ s = ut + \frac{1}{2} a t^2 \] Substitute the values: \[ s = (20 \times 5) + \frac{1}{2} (-4) (5)^2 \] \[ s = 100 - 50 = 50 \, \text{meters} \] **Verification:** - Alternative method: Average velocity = \(\frac{u + v}{2} = \frac{20 + 0}{2} = 10 \, \text{m/s}\). Distance = Average velocity × time = \(10 \times 5 = 50 \, \text{meters}\). - Both methods confirm the distance is **50 meters**. **Final Answer:** The car’s acceleration is **\(-4 \, \text{m/s}^2\)** (deceleration), and the distance traveled during deceleration is **50 meters**. 1 Show Full Solution Final Answer Understanding the Solution Share Solution
