the speed of the train is reduce uniformly from 15m/s to 7m/s while travelling a distance of 90m.
a) compute the acceleration
b) how much further will the train travel before coming to rest provided the acceleration remain constant.
a) The average speed during the interval is (1/2)(15+7) = 11 m/s
The time required to travel 90 m while decelerating from 15 to 7 m/s is
t = 90/11 = 8.18 s
The acceleration rate during that interval is
a = (7 - 15)/8.18 = -0.978 m/s^2
b) To stop, the additional time required is (7 m/s)/(0.978 m/s^2) = 7.158 s
Multiply that by the average speed during that interval, 3.5 m/s.
25.06 m/s
well worked,but you should be explaining in details about you find the answers
Great!
To compute the acceleration, we can use the formula:
a = (vf - vi) / t
where:
a = acceleration
vf = final velocity
vi = initial velocity
t = time
In this case, the initial velocity (vi) is 15 m/s, the final velocity (vf) is 7 m/s, and the time (t) is not given. However, we can find the time using the formula:
t = d / v_avg
where:
d = distance traveled
v_avg = average velocity
In this case, the distance traveled (d) is 90 m. The average velocity (v_avg) can be found by taking the average of the initial and final velocities:
v_avg = (vi + vf) / 2
Now, let's calculate the acceleration (a):
v_avg = (15 + 7) / 2 = 11 m/s
t = 90 / 11 = 8.18 s
a = (7 - 15) / 8.18 = -0.975 m/s² (note that it is negative because the velocity is decreasing)
a) The acceleration of the train is approximately -0.975 m/s².
To calculate the distance the train will travel before coming to rest, we can use the formula:
d = (vf^2 - vi^2) / (2a)
Since the final velocity (vf) is 0 m/s (as the train comes to rest), the formula simplifies to:
d = -vi^2 / (2a)
Plugging in the values:
d = -15^2 / (2 * -0.975) = 112.82 m
b) The train will travel approximately 112.82 meters further before coming to rest, provided the acceleration remains constant.