A body moving with uniform acceleration has a velocity of 9.9 cm/s when its x coordinate is 1.28 cm. If its x coordinate 1.3 s later is -9.1 cm, what is the magnitude of its acceleration? Answer in units of cm/s2.
Start t=0 at the X = 1.28 cm location
X = (a/2)t^2 + 9.9 t + 1.28
-9.1 = (a/2)(1.3)^2 + 9.9*1.3 + 1.28
0.845 a = -9.1 -1.28 -12.87
a = -27.45 cm/s^2
To find the magnitude of the acceleration, we need to solve this problem in two steps:
Step 1: Find the change in velocity (Δv)
Step 2: Use the kinematic equation to find the magnitude of acceleration (a)
Step 1: Find the change in velocity (Δv)
We are given:
Initial velocity (u) = 9.9 cm/s
Final velocity (v) = unknown
Time (t) = 1.3 seconds
We know that acceleration (a) is constant, so we can use the equation:
v = u + at
Rearranging the equation to solve for the final velocity (v):
v = u + at
v = 9.9 cm/s + a * 1.3 s
v = 9.9 cm/s + 1.3a
Step 2: Use the kinematic equation to find the magnitude of acceleration (a)
We are given:
Initial x-coordinate (x₁) = 1.28 cm
Final x-coordinate (x₂) = -9.1 cm
Time (t) = 1.3 seconds
We know that displacement (Δx) can be calculated using the equation:
Δx = ut + (1/2)at²
Since the body starts at x₁ and ends up at x₂, we can write:
x₂ - x₁ = ut + (1/2)at²
-9.1 cm - 1.28 cm = 0 + (1/2)a * (1.3 s)²
-10.38 cm = 0.65a * 1.69 s²
-10.38 cm = 1.0975a
Now, we have two equations:
v = 9.9 cm/s + 1.3a
-10.38 cm = 1.0975a
Solving these equations simultaneously will give us the magnitude of acceleration (a).
Substitute the value of v from the first equation into the second equation:
-10.38 cm = 1.0975(9.9 cm/s + 1.3a)
-10.38 cm = 10.8225 cm/s + 1.42275a
1.42275a = -10.38 cm - 10.8225 cm/s
1.42275a = -21.2025 cm/s
a = -14.9 cm/s²
Therefore, the magnitude of the acceleration is 14.9 cm/s².