The position of a particle as it moves along the x axis is given by x = 15eƒ{2t m, where t is in s. What is the acceleration of the particle at t = 1.0 s?
4.
To find the acceleration of the particle at t = 1.0 s, we need to take the second derivative of the position function with respect to time (t). The position function is given as x = 15e^(2t) m.
The first derivative of the position function gives us the velocity function, v(t):
v(t) = dx/dt = d/dt (15e^(2t))
To find this derivative, we can use the chain rule. The derivative of e^(2t) is 2e^(2t), and the derivative of 15 is 0 since it is a constant. Therefore, the velocity function is:
v(t) = 30e^(2t)
Now, to find the acceleration at t = 1.0 s, we need to take the derivative of the velocity function with respect to t. Let's call this derivative a(t):
a(t) = dv/dt = d/dt (30e^(2t))
Again, we can use the chain rule to differentiate e^(2t), which gives us 2e^(2t). Differentiating the constant 30 results in 0. Therefore, the acceleration function is:
a(t) = 60e^(2t)
Now, substitute t = 1.0 into the acceleration function to find the acceleration at t = 1.0 s:
a(1.0) = 60e^(2*1.0)
a(1.0) = 60e^2
Calculating this value, we get:
a(1.0) ≈ 164.87 m/s^2
Therefore, the acceleration of the particle at t = 1.0 s is approximately 164.87 m/s^2.