A student starts from rest from his house and accelerates uniformly at the rate of 2m/sec^2 for 30 sec then he apply break for 45 sec and come to rest at his school.how far is his school
d = 0.5a*t^2
d1 = 1*30^2 = 900 m. = Distance traveled in 30 s.
V = at = 2*30 = 60 m/s = Velocity reached in 30 s.
a=(V-Vo)/t = (0-60) / 45 = -1.33 m/s^2.
V^2 = Vo^2 + 2a*d
0 = 60^2 - 2.66d
2.666d = 3600
d2 = 1350 m.
d1 + d2 = 900 + 1350 = 2250 m. =
Distance from house to school.
To find the distance to the school, we need to determine the distance traveled during the acceleration phase and the distance traveled during the deceleration phase.
During uniform acceleration, we can use the equation:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity
a = acceleration
t = time
In this case, the student starts from rest, so the initial velocity (u) is 0 m/s. The acceleration (a) is 2 m/s^2, and the time (t) is 30 seconds.
Plugging these values into the equation, we get:
s1 = (0)(30) + (1/2)(2)(30^2)
s1 = 0 + (1/2)(2)(900)
s1 = 0 + 1800
s1 = 1800 meters
Therefore, the distance traveled during the acceleration phase is 1800 meters.
During the deceleration phase, the student applies the brakes for 45 seconds until he comes to rest. Since the brakes are applied uniformly, we can use the same equation to find the distance traveled.
In this case, the initial velocity (u) is the final velocity from the accelerating phase, which is also 0 m/s. The acceleration (a) is the same as before, 2 m/s^2, but the time (t) is 45 seconds.
Plugging these values into the equation, we get:
s2 = (0)(45) + (1/2)(2)(45^2)
s2 = 0 + (1/2)(2)(2025)
s2 = 0 + 4050
s2 = 4050 meters
Therefore, the distance traveled during the deceleration phase is 4050 meters.
To find the total distance to the school, we add the distances traveled during the acceleration and deceleration phases:
Total distance = s1 + s2
Total distance = 1800 + 4050
Total distance = 5850 meters
Therefore, the school is 5850 meters away from the student's house.