a particle moves on a circular path of 20 ft radius so that its arc distance from a fixed point of the path is given by s= 4 t^2-10t where a is in feet and t is in seconds. compute the total acceleration at the end of 2 sec.
sda
1eqweqeqw
To find the total acceleration at the end of 2 seconds, we need to differentiate the equation for s with respect to time (t) twice.
Given: s = 4t^2 - 10t
Step 1: Find the first derivative of s with respect to t (ds/dt):
ds/dt = d/dt (4t^2 - 10t)
= 8t - 10
The first derivative gives us the instantaneous velocity of the particle.
Step 2: Find the second derivative of s with respect to t (d^2s/dt^2):
d^2s/dt^2 = d/dt (8t - 10)
= 8
The second derivative gives us the instantaneous acceleration of the particle.
Step 3: Substitute t = 2 into the second derivative equation to find the acceleration at the end of 2 seconds:
d^2s/dt^2 = 8 (when t = 2)
= 8 ft/s^2
Therefore, the total acceleration at the end of 2 seconds is 8 ft/s^2.
To compute the total acceleration at the end of 2 seconds, we need to find two components of acceleration: the tangential acceleration and the radial acceleration.
First, let's find the tangential acceleration. Tangential acceleration (at) represents the rate of change of speed with respect to time. It is given by the second derivative of the arc length function (s) with respect to time (t).
The given arc length function is s = 4t² - 10t. To find the derivative of s with respect to t, we can apply the power rule for derivatives:
s' = d(s)/d(t) = d(4t² - 10t)/d(t) = 8t - 10.
Now, we can find the second derivative by differentiating s' with respect to t:
s'' = d(s')/d(t) = d(8t - 10)/d(t) = 8.
So, the tangential acceleration (at) is constant and equal to 8 ft/s².
Next, let's find the radial acceleration. Radial acceleration (ar) represents the acceleration pointing towards the center of the circle.
The radial acceleration can be determined using the formula:
ar = (v²) / r,
where v represents the particle's instantaneous speed and r is the radius of the circular path.
To find v, we need to find the derivative of s' with respect to t:
v = d(s')/d(t) = d(8t - 10)/d(t) = 8.
Now, we can substitute the values of v and r (20 ft) into the formula for radial acceleration:
ar = (8²) / 20 = 64 / 20 = 3.2 ft/s².
Finally, to compute the total acceleration (atotal) at the end of 2 seconds, we can use the Pythagorean theorem because the tangential and radial accelerations are perpendicular to each other:
atotal = √(at² + ar²) = √(8² + 3.2²) = √(64 + 10.24) = √(74.24) ≈ 8.62 ft/s².
Therefore, the total acceleration at the end of 2 seconds is approximately 8.62 ft/s².