a particle starts from rest and moves along a straight line with an acceleration equal to 24-15√t where distance and time are measured in terms of feet and seconds respectively. with what acceleration will the particle return to the starting point?
To find the acceleration at which the particle returns to the starting point, we need to determine the time when the particle will reach back to its original position.
Given that the particle starts from rest, we know its initial velocity, v₀, is zero.
The equation of motion for the particle with constant acceleration is:
v = v₀ + at
We can integrate the given acceleration equation to find the velocity function for the particle:
a = 24 - 15√t
v = ∫(24 - 15√t) dt
To find the time at which the particle reaches back to its starting point, we need to solve for t when the displacement, s, is zero. The equation can be written as:
0 = ∫(24 - 15√t) dt
To solve this integral, we will break it down into two parts:
∫24 dt - ∫15√t dt
Integrating the first part is easy:
∫24 dt = 24t
Now, let's focus on the second part:
∫15√t dt
Using the power rule of integration, we can integrate this as:
∫15√t dt = 15 * (2/3) * t^(3/2) + C
where C is the constant of integration.
So, the total integral is:
0 = 24t - 15 * (2/3) * t^(3/2) + C
Now, to find the time at which the particle returns to the starting point (when s = 0), we can solve this equation for t.
0 = 24t - 15 * (2/3) * t^(3/2) + C
Since we know the particle starts from rest, the constant of integration, C, will be zero.
0 = 24t - 15 * (2/3) * t^(3/2)
We can simplify this equation to:
0 = 8t - 10 * t^(3/2)
To solve this equation, we can factor out t and rearrange:
0 = t * (8 - 10√t)
From this equation, we can see that the particle will return to the starting point at t = 0 (the initial position) or when the term inside the parentheses becomes zero:
8 - 10√t = 0
10√t = 8
√t = 8/10
Squaring both sides:
t = (8/10)^2
Simplifying the equation further:
t = 64/100
t = 0.64 seconds
Now that we have the time when the particle returns to the starting point, we can find the acceleration by substituting this time value into the given acceleration equation:
a = 24 - 15√t
a = 24 - 15√(0.64)
a ≈ 24 - 15(0.8)
a ≈ 24 - 12
a ≈ 12 ft/s²
Therefore, the acceleration at which the particle returns to the starting point is approximately 12 ft/s².