an object moving on the x axis with a constant acceleration increases its x coordinate by 108 m in a time of 9.5 s and has a velocity of 20 m/s at the end of this time.
Determine the acceleration of the object during this motion?
Well, if the object is moving on the x-axis with constant acceleration and it increases its x-coordinate by 108 m in a time of 9.5 s, then it must be participating in a very intense game of "extreme leapfrog" with some invisible friends!
But let's get serious for a moment. We can use the kinematic equation to determine the acceleration of the object. The equation is:
x = x₀ + v₀t + 0.5at²
Where:
- x is the final position (108 m)
- x₀ is the initial position (0 m)
- v₀ is the initial velocity (unknown)
- t is the time (9.5 s)
- a is the acceleration (unknown)
Now, the object has a velocity of 20 m/s at the end of this time. We can use this information to find the value of the initial velocity, v₀.
v = v₀ + at
Substituting the values in, we get:
20 m/s = v₀ + a * 9.5 s
Now, we have two equations and two unknowns. Let's solve these simultaneously and see if they reveal any surprising secrets!
To determine the acceleration of the object, we can use the equations of motion. Here's how you can calculate it step-by-step:
Step 1: Write down the given information:
- Change in position (Δx) = 108 m
- Time (t) = 9.5 s
- Final velocity (v) = 20 m/s
Step 2: Use the equation of accelerated motion to find the acceleration (a):
Δx = v_initial * t + 0.5 * a * t^2
Since the object is moving with a constant acceleration, the initial velocity (v_initial) can be assumed to be zero.
Substituting the values, the equation becomes:
108 = 0 * 9.5 + 0.5 * a * (9.5)^2
Step 3: Simplify the equation:
108 = 0 + 0.5 * a * 90.25
Step 4: Solve for acceleration (a):
108 = 45.125 * a
Divide both sides by 45.125:
a = 108 / 45.125
Step 5: Calculate the acceleration:
a ≈ 2.39 m/s^2
Therefore, the acceleration of the object during this motion is approximately 2.39 m/s^2.
To determine the acceleration of the object during its motion, we can use the kinematic equation:
Δx = v₀t + (1/2)at²
Where:
Δx is the change in position (in this case, 108 m)
v₀ is the initial velocity (in this case, 0 m/s)
t is the time interval (in this case, 9.5 s)
a is the acceleration we want to find
Since the object has a velocity of 20 m/s at the end of the time interval, we can add this information to the equation:
Δx = v₀t + (1/2)at²
108 m = 0 m/s * 9.5 s + (1/2) * a * (9.5 s)²
108 m = 0 + (1/2) * a * (90.25 s²)
108 m = 45.125a * s²
a = 108 m / (45.125 s²)
a ≈ 2.395 m/s²
Therefore, the acceleration of the object during this motion is approximately 2.395 m/s².
v = Vi + a t
20 = Vi + 9.5 a
d = Vi t + (1/2) a t^2
108 = (20-9.5 a) (9.5) + (1/2) a (9.5)^2
solve for a