Mrs bautista accelerates her bicycle uniformly along a straight path from 10.5 m/s to rest in 5.2s.
A. What is her acceleration?
B. What is her displacement during this time interval?
a = change in velocity / change in time
= ( 0 - 10.5) / 5.2 m/s^2
x = (1/2) a t^2
= (1/2) * a from above * (5.2) * (5.2)
by the way part B is also the average speed times the time
x = 5.25 * 5.2
To find the answers, we will use the equations of motion for uniformly accelerated motion.
A. To find the acceleration (a), we can use the equation:
v = u + at
Where:
v = final velocity = 0 m/s (because she comes to rest)
u = initial velocity = 10.5 m/s
t = time taken = 5.2 s
By rearranging the equation, we can solve for acceleration (a):
0 = 10.5 + a * 5.2
Subtracting 10.5 from both sides:
-10.5 = 5.2a
Dividing both sides by 5.2:
a = -10.5 / 5.2 = -2.019 m/s^2 (rounded to three decimal places)
Therefore, Mrs. Bautista's acceleration is approximately -2.019 m/s^2 (negative because the bicycle is decelerating).
B. To find the displacement, we can use another equation of motion:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity = 10.5 m/s
a = acceleration = -2.019 m/s^2
t = time taken = 5.2 s
Plugging in the values:
s = 10.5 * 5.2 + (1/2) * (-2.019) * (5.2)^2
Simplifying the equation:
s = 54.6 - 26.7
s = 27.9 meters
Therefore, Mrs. Bautista's displacement during this time interval is 27.9 meters.