HOME
We are at the home screen. The user has just typed "HOME". This might be a command to go back to the home screen, or it could be a typo, or it might be the start of a message.
Since the context is unclear, I'll respond by acknowledging the home screen and offering assistance.
Response:
"Welcome to the home screen! How can I assist you today?"
Welcome to the home screen! 馃彔
Here鈥檚 what you can do next:
馃殌 Quick Actions:
Explore Features:
鉁?Check your tasks
馃搳 View analytics
馃摤 Manage messages
鈿欙笍 Settings
Need Help?
Type help for commands.
Ask anything like "What can you do?"
馃挕 Try Examples:
"Show my tasks"
"Set a reminder for tomorrow"
"What's the weather today?"
Just let me know what you鈥檇 like to do! 馃槉
Okay, so I need to figure out how to solve this problem where a car accelerates from rest with a constant acceleration of 2 m/s虏. The question is asking for the time it takes to reach a speed of 25 m/s. Alright, let's start by recalling some basic physics equations related to motion.
First, since the car starts from rest, the initial velocity, u, is 0 m/s. The acceleration, a, is given as 2 m/s虏. The final velocity, v, we want is 25 m/s. We need to find the time, t.
I remember that one of the kinematic equations for uniformly accelerated motion is v = u + at. That seems perfect here because we have all the variables except time. Let me write that down:
v = u + a*t
Plugging in the known values:
25 m/s = 0 m/s + 2 m/s虏 * t
So, simplifying that equation:
25 = 2t
https://www.18122.com/url/usdt/usdt0379.jpg
To solve for t, divide both sides by 2:
t = 25 / 2
Calculating that gives t = 12.5 seconds.
Wait, let me double-check. If the car accelerates at 2 m/s every second, then after 1 second it's going 2 m/s, after 2 seconds 4 m/s, and so on. So, to reach 25 m/s, dividing 25 by 2 should indeed give 12.5 seconds. That makes sense.
Is there another way to verify this? Maybe using another kinematic equation, like the one involving distance? But since we don't have distance, maybe not necessary here. Alternatively, think about the units. Acceleration is in m/s虏, time in seconds, so multiplying acceleration by time gives m/s, which matches velocity. The units check out.
Hmm, seems straightforward. Maybe I can also think about the graph of velocity vs. time. Since it's a constant acceleration, the graph would be a straight line starting from (0,0) with a slope of 2. The equation of that line is v = 2t. Setting v to 25, solve for t, same result.
I don't see any mistakes here. The answer should be 12.5 seconds.
To determine the time it takes for a car accelerating from rest with a constant acceleration of (2 , \text{m/s}^2) to reach a speed of (25 , \text{m/s}), we use the kinematic equation:
[
v = u + a t
]
Where:
(v = 25 , \text{m/s}) (final velocity),
(u = 0 , \text{m/s}) (initial velocity, since the car starts from rest),
(a = 2 , \text{m/s}^2) (acceleration),
(t) is the time to solve for.
Plugging in the values:
[
25 = 0 + 2t
]
Solving for (t):
[
t = \frac{25}{2} = 12.5 , \text{seconds}
]
Answer: The car takes (\boxed{12.5}) seconds to reach a speed of (25 , \text{m/s}).
页:
[1]