To solve these problems, we will use the equations of motion for uniformly accelerated motion.
1. To find the acceleration and stopping distance of the train, we can use the equation:
v = u + at
Where:
v = final velocity (0 m/s, because the train comes to a stop)
u = initial velocity (30 m/s)
a = acceleration (unknown)
t = time taken to stop (44 seconds)
Rearranging the equation to solve for acceleration (a), we have:
0 = 30 + a * 44
Solving for 'a', we get:
a = (0 - 30) / 44
a ≈ -0.68 m/s²
This negative value indicates deceleration (slowing down) of the train.
To find the stopping distance, we can use the equation:
s = ut + 1/2 * a * t²
Plugging in the values we know:
s = 30 * 44 + 1/2 * (-0.68) * (44)²
s ≈ 660 meters
Therefore, the acceleration is approximately -0.68 m/s², and the stopping distance is approximately 660 meters.
2. To find the acceleration and the time taken for the car, we can use the following equation:
v² = u² + 2as
Where:
v = final velocity (20 m/s)
u = initial velocity (6 m/s)
a = acceleration (unknown)
s = distance covered (70 meters)
Rearranging the equation to solve for acceleration (a), we have:
a = (v² - u²) / (2s)
a = (20² - 6²) / (2 * 70)
a ≈ 1.72 m/s²
Hence, the acceleration is approximately 1.72 m/s².
To find the time taken (t), we can use the equation:
v = u + at
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a
t = (20 - 6) / 1.72
t ≈ 8.14 seconds
Therefore, the acceleration is approximately 1.72 m/s², and the time taken is approximately 8.14 seconds.