A vehicle starts from rest and accelerates with a constant acceleration, covering 200m in 5s. it then continues at a constant velocity for 300m before decelerating with a constant 2m/s^2 deceleration to a stop. what is the maximum speed attained, total time traveled, total distance covered?

Question

A vehicle starts from rest and accelerates with a constant acceleration, covering 200m in 5s. it then continues at a constant velocity for 300m before decelerating with a constant 2m/s^2 deceleration to a stop. what is the maximum speed attained, total time traveled, total distance covered?

Answer 1

200 meters in 5 seconds

average speed = 200/5 = 40 m/s
so speed is 80 m/s at 200 m
it slows down after that so 80 is your answer
goes at 80 for 300 m
time = 300/80 = 30/8 = 15/4 seconds between 200 and 500
now from 500 m and 80 m/s, brakes down to zero at -2 m/s^2
0 = 80 - 2 t
t = 40 seconds
so time total = 5 + 15/4 + 40
average speed during braking phase = 40
so from 500 m to stop it covers
40 * 40 =1600 meters
500 + 1600 = 2100 meters total

Answer 2

To solve this problem, we'll break it down into three parts: the first part where the vehicle accelerates, the second part where it travels at a constant velocity, and the third part where it decelerates.

Part 1: Acceleration
We are given that the vehicle covers a distance of 200m in 5s with constant acceleration. Using the formula for distance covered during uniformly accelerated motion:

𝑑 = 𝑣₀𝑡 + 0.5𝑎𝑡²

where 𝑑 is the distance traveled, 𝑣₀ is the initial velocity, 𝑡 is the time, and 𝑎 is the acceleration.

As the vehicle starts from rest, 𝑣₀ = 0.

Since we know the distance (200m), time (5s), and we need to find the acceleration, we can rearrange the formula to solve for 𝑎.

200 = 0.5𝑎(5)²
200 = 12.5𝑎
𝑎 = 16 m/s²

Part 2: Constant Velocity
The vehicle travels at a constant velocity for 300m. This means there is no acceleration involved. Therefore, the time taken can be calculated using the formula:

𝑡 = 𝑑 / 𝑣

where 𝑡 is the time, 𝑑 is the distance, and 𝑣 is the velocity.

𝑡 = 300 / 𝑣

Part 3: Deceleration
The vehicle decelerates with a constant deceleration of 2 m/s² until it comes to a stop. Similar to the acceleration, we can use the formula for distance covered during uniformly accelerated motion:

𝑑 = 𝑣₀𝑡 + 0.5𝑎𝑡²

As the final velocity is 0 (since it comes to a stop), and the deceleration is given (-2 m/s²), we can rearrange the formula accordingly:

0 = 𝑣² + 2𝑎𝑑

𝑣² = -2𝑎𝑑
𝑣 = -√(2𝑎𝑑)

Here, 𝑣 is the maximum speed attained, 𝑎 is the deceleration, and 𝑑 is the distance covered.

Now, let's calculate the answers:

Part 1:
𝑎 = 16 m/s² (acceleration)
𝑑 = 200 m (distance)
𝑡 = 5 s (time)

Part 2:
𝑑 = 300 m (distance)
𝑡 = 300 / 𝑣 (time)

Part 3:
𝑎 = -2 m/s² (deceleration)
𝑑 = 300 m (distance)

Finally, we can find the total time traveled, total distance covered, and the maximum speed attained by summing up the times and distances in each part.