An object moving on the x axis with a constant acceleration increases its x coordinate by 161 m in a time of 12.5 s and has a velocity of 20 m/s at the end of this time.
Determine the acceleration of the object
during this motion.
Answer in units of m/s2.
-2 X (161-20 X 12.5) / 12.5^2
=1.139
Well, first of all, let's commend this object for its dedication to increasing its x coordinate by 161 m in just 12.5 seconds. That's some serious commitment to forward motion!
Now, let's get down to business. To determine the acceleration of the object, we can use the kinematic equation:
x = x0 + v0t + (1/2)at^2
Here, x is the final position (161 m), x0 is the initial position (which we don't know yet), v0 is the initial velocity (0 m/s because the object starts from rest), t is the time (12.5 s), and a is the acceleration (what we're trying to find).
Since the object has a velocity of 20 m/s at the end of the 12.5-second time period, we can use that information to find the initial position:
v = v0 + at
20 m/s = 0 m/s + a(12.5 s)
20 m/s = 12.5a m/s^2
Solving for a, we get:
a = 20 m/s / 12.5 s
a ≈ 1.6 m/s^2
So, the acceleration of this object during its motion is approximately 1.6 m/s^2. It's like the object was saying, "I'm going to accelerate just enough to make my x coordinate increase by 161 m in 12.5 seconds. No more, no less!"
To determine the acceleration of the object, we can use the kinematic equation:
x = (1/2)at^2 + v₀t + x₀
where:
- x is the displacement (161 m)
- a is the acceleration (unknown)
- t is the time (12.5 s)
- v₀ is the initial velocity (unknown)
- x₀ is the initial position (assumed to be zero, since the problem does not specify)
We are given that v = 20 m/s at t = 12.5 s. This means the object's velocity at the end of this time is 20 m/s. Therefore:
v = v₀ + at
Substituting the given values:
20 = v₀ + a * 12.5
To find the acceleration, we need to solve this equation for a. However, we need the value of v₀ to proceed. To find v₀, we can use the displacement equation:
x = (1/2)at² + v₀t + x₀
Substituting the given values:
161 = (1/2) * a * (12.5)^2 + v₀ * 12.5 + 0
Converting the equation to solve for v₀:
⇒ 20 * 12.5 = (1/2) * a * (12.5)^2 + v₀ * 12.5
⇒ 250 = (1/2) * a * (12.5)^2 + v₀ * 12.5
Now, we have two equations:
20 = v₀ + a * 12.5
250 = (1/2) * a * (12.5)^2 + v₀ * 12.5
We can solve this system of equations to find the values of a and v₀.
To determine the acceleration of the object during this motion, we can use the following kinematic equation:
x = x0 + v0t + (1/2)at^2
where:
x is the final position (161 m)
x0 is the initial position (0 m)
v0 is the initial velocity (unknown)
t is the time (12.5 s)
a is the acceleration (unknown)
We are given the final position (161 m), time (12.5 s), and final velocity (20 m/s).
Using the kinematic equation, we can rearrange it to solve for acceleration (a):
x = x0 + v0t + (1/2)at^2
161 = 0 + v0(12.5) + (1/2)a(12.5)^2
161 = 12.5v0 + 78.125a
We also know that at the end of 12.5 seconds, the object has a velocity of 20 m/s. So we can write:
v = v0 + at
20 = v0 + a(12.5)
20 = v0 + 12.5a
Now we have a system of equations:
161 = 12.5v0 + 78.125a
20 = v0 + 12.5a
To solve this system of equations, we can use the method of substitution. Rearrange the second equation to solve for v0:
v0 = 20 - 12.5a
Substitute this value of v0 in the first equation:
161 = 12.5(20 - 12.5a) + 78.125a
Simplify the equation:
161 = 250 - 156.25a + 78.125a
Combine like terms:
161 = 250 - 78.125a
Rearrange the equation:
78.125a = 250 - 161
Evaluate the right side:
78.125a = 89
Divide both sides by 78.125 to solve for a:
a = 89 / 78.125
Evaluate the expression:
a ≈ 1.14 m/s^2
Therefore, the acceleration of the object during this motion is approximately 1.14 m/s^2.