a bus moving at a speed of 20m/s begins to slow at a constant rate. Find how far it goes before stopping?
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To find the distance the bus goes before stopping, we need to determine the time it takes to stop. We can use the equation of motion:
v = u + at
Where:
v = final velocity (which is 0 m/s, since the bus stops)
u = initial velocity (20 m/s)
a = acceleration (negative, since the bus is slowing down)
t = time
Since the final velocity is 0 and the initial velocity is 20 m/s, we have:
0 = 20 + a * t
Rearranging the equation, we get:
t = -20/a
Now, we need to find the acceleration. The question states that the bus slows down at a constant rate, so we can assume that the acceleration is constant. Let's denote the acceleration as "a".
Now, to find "a", we use another equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s)
u = initial velocity (20 m/s)
a = acceleration
s = distance
Substituting the known values:
0 = (20)^2 + 2*a*s
400 = 2*a*s
Dividing both sides by 2a, we get:
200 = a*s
Simplifying further:
s = 200/a
Now we can substitute the value of "t" in terms of "a" into our equation for "s":
s = 200/(-20/a)
Simplifying:
s = -200a/20
s = -10a
Therefore, the distance the bus goes before stopping is -10a meters. However, without knowing the value of the acceleration, we cannot determine the exact distance.