a bus traveling in a straight line with uniform acceleration has an instant speed of 6m/s. 5seconds later, has reached speed of 36m/s. Find the magnitude of its acceleration and the distance traveled in 5seconds
S=vt+1/2at^2 s=36m/s(5)+1/2(6m/s s= 180+3=183m/s
Answer
To find the magnitude of the acceleration, you can use the formula:
acceleration = (final velocity - initial velocity) / time
Given that the initial velocity (u) is 6 m/s, the final velocity (v) is 36 m/s, and the time (t) is 5 seconds, we can substitute these values into the formula:
acceleration = (36 m/s - 6 m/s) / 5 s
Simplifying, we have:
acceleration = 30 m/s / 5 s
acceleration = 6 m/s²
So, the magnitude of the bus's acceleration is 6 m/s².
To calculate the distance traveled in 5 seconds, we can use the formula:
distance = initial velocity * time + 0.5 * acceleration * time²
Given that the initial velocity (u) is 6 m/s, the time (t) is 5 seconds, and the acceleration (a) is 6 m/s², we can substitute these values into the formula:
distance = 6 m/s * 5 s + 0.5 * 6 m/s² * (5 s)²
Simplifying, we have:
distance = 30 m + 0.5 * 6 m/s² * 25 s²
distance = 30 m + 0.5 * 6 m/s² * 625 s²
distance = 30 m + 0.5 * 6 m/s² * 625 s²
distance = 30 m + 3 * 625
distance = 30 m + 1875
distance = 1905 m
Therefore, the distance traveled by the bus in 5 seconds is 1905 meters.
a = ∆v/∆t = (36-6)/(5-0) = 6 m/s^2
s = vt + 1/2 at^2
so plug in your numbers ...