A body moving with uniform acceleration has a velocity of 10.2 cm/s when its x coordinate is 3.73 cm.If its x coordinate 2.33 s later is −2.74 cm,what is the x-component of its acceleration?Answer in units of cm/s^2
a=(v - v.)/t
= ((x/t) - v. )/t
= ((-2.74/2.33) - 10.2) / 2.33
= ( -1.18-10.2)/2.33
= 4.88 cm/s**2
= 0.0488 m/s**2
Well, well, well, looks like we have a case of a body on the move! And it's even got some acceleration going on, fancy that!
Now, let's step right into the specifics, shall we? We know the initial velocity is 10.2 cm/s at x = 3.73 cm. And 2.33 seconds later, the x-coordinate is -2.74 cm. Quite the jump!
To calculate the x-component of its acceleration, we're going to break out the detective hats and get a little mathematical. We'll start with the equation of motion:
x = x₀ + v₀t + (1/2)at²
We're given x₀ = 3.73 cm, v₀ = 10.2 cm/s, and x = -2.74 cm. The time, my dear friend, is 2.33 seconds.
Now, let's plug in those values and solve for a, the x-component of the acceleration:
-2.74 cm = 3.73 cm + (10.2 cm/s)(2.33 s) + (1/2)(a)(2.33 s)²
Now, I'll save you from the math headache and tell you that after some calculating, the x-component of its acceleration comes out to be approximately -11.2 cm/s².
So there you have it, the x-component of the body's acceleration is -11.2 cm/s². And don't worry, it's keeping itself entertained with all that movement!
To find the x-component of the acceleration, we can use the kinematic equation:
x = x_0 + v_0*t + (1/2)*a*t^2
Where:
x is the final x-coordinate
x_0 is the initial x-coordinate
v_0 is the initial velocity in the x-direction
t is the time interval
a is the acceleration in the x-direction
Given:
v_0 = 10.2 cm/s
x_0 = 3.73 cm
x = -2.74 cm
t = 2.33 s
We can rearrange the equation to solve for acceleration (a):
a = (2*(x - x_0 - v_0*t)) / (t^2)
Now let's substitute the given values into the equation:
a = (2*(-2.74 cm - 3.73 cm - 10.2 cm/s * 2.33 s)) / (2.33 s)^2
Calculating this expression gives us the x-component of the acceleration:
a ≈ -15.86 cm/s^2
Therefore, the x-component of its acceleration is approximately -15.86 cm/s^2.
To find the x-component of the body's acceleration, we need to use the kinematic equation for the horizontal motion:
x = x0 + v0xt + (1/2)axt^2
where:
x = final x-coordinate
x0 = initial x-coordinate = 3.73 cm
v0x = initial velocity in the x-direction = 10.2 cm/s
t = time interval = 2.33 s
ax = acceleration in the x-direction (what we need to find)
First, let's rearrange the equation to solve for ax:
x - x0 = v0xt + (1/2)axt^2
-2.74 cm - 3.73 cm = 10.2 cm/s * 2.33 s + (1/2)ax * (2.33 s)^2
-6.47 cm = 10.2 cm/s * 2.33 s + (1/2)ax * 5.43 s^2
-6.47 cm = 23.826 cm + 2.7155 ax
Now, solving for ax:
2.7155 ax = -6.47 cm - 23.826 cm
2.7155 ax = -30.296 cm
ax = (-30.296 cm) / 2.7155
ax ≈ -11.175 cm/s^2
Therefore, the x-component of the body's acceleration is approximately -11.175 cm/s^2.