Suppose that a particle moves along a line so that its velocity v at time t is given by this piecewise function:
v(t)=5t if 0≤t<1
v(t)=6((t)^(1/2))-(1/t) if 1≤t
where t is in seconds and v is in centimeters per second (cm/s).
Estimate the time(s) at which the particle is 4 cm from its starting position
Assuming an initial position of zero,
s(t) = 5/2 t^2 for 0<=t<1
so, at t=1, s = 5/2
Now, using the 2nd function,
s(t) = 5/2 + 4t^(3/2) - log(t)
solve that for s(t) = 4
To estimate the time(s) at which the particle is 4 cm from its starting position, we need to find the time(s) when the displacement is 4 cm.
The displacement of the particle can be found by integrating its velocity function.
For the first interval, where 0 ≤ t < 1, the velocity function is v(t) = 5t.
Integrating v(t) with respect to t, we have:
s(t) = ∫ v(t) dt
= ∫ 5t dt
= (5/2) t^2 + C
Since we are given that the particle starts at t = 0, its initial position is s(0) = 0 cm.
So, the equation for the displacement during this interval is:
s(t) = (5/2) t^2
For the second interval, where t ≥ 1, the velocity function is v(t) = 6√t - 1/t.
Integrating v(t) with respect to t, we have:
s(t) = ∫ (6√t - 1/t) dt
= (4/3) t^(3/2) - ln|t| + C
Since we are given that the particle starts at t = 0, its initial position is s(0) = 0 cm.
So, the equation for the displacement during this interval is:
s(t) = (4/3) t^(3/2) - ln|t|
We need to find the time(s) when the displacement is 4 cm, so we can set up the following equation for each interval and solve for t:
For the first interval:
(5/2) t^2 = 4
Solving this quadratic equation, we get:
t^2 = (8/5)
t = ±√(8/5)
Since time cannot be negative, we take the positive value, t = √(8/5) ≈ 1.264 s
For the second interval:
(4/3) t^(3/2) - ln|t| = 4
We can solve this equation numerically using a graphing calculator or computer software to find the value(s) of t. Depending on the accuracy required, there may be multiple solutions.
Therefore, the estimated time(s) at which the particle is 4 cm from its starting position are approximately t = 1.264 seconds for the first interval and the solution(s) for the second interval.
To find the time(s) when the particle is 4 cm from its starting position, we need to determine when the displacement of the particle is equal to 4 cm. Displacement is calculated by integrating the velocity function.
Let's split the problem into two cases based on the given piecewise function:
Case 1: 0 ≤ t < 1
In this range, the particle's velocity is given by v(t) = 5t.
To find the displacement, we integrate the velocity function within the given time range:
∫(5t) dt = (5/2)t^2 + C
The constant of integration C is determined by the initial condition, which is the starting position of the particle. Since the problem doesn't provide a specific starting position, we'll assume it's zero.
Therefore, the displacement function within this time range is D1(t) = (5/2)t^2.
To find when the particle is 4 cm from its starting position, we set the displacement function equal to 4:
(5/2)t^2 = 4
t^2 = (4 * 2) / 5
t^2 = 8/5
t = √(8/5)
Case 2: t ≥ 1
In this range, the particle's velocity is given by v(t) = 6√(t) - (1/t).
To find the displacement, we integrate the velocity function within the given time range:
∫(6√(t) - (1/t)) dt = 2t^(3/2) - ln(t) + C
Again, the constant of integration C is determined by the initial condition. Since we assume a starting position of zero, C becomes zero.
Therefore, the displacement function within this time range is D2(t) = 2t^(3/2) - ln(t).
To find when the particle is 4 cm from its starting position, we set the displacement function equal to 4:
2t^(3/2) - ln(t) = 4
2t^(3/2) = 4 + ln(t)
t^(3/2) = 2 + (1/2)ln(t)
Solving this equation algebraically is not straightforward, but we can use numerical methods, such as iteration or graphing, to approximate the solution.
In summary,
For 0 ≤ t < 1: t ≈ √(8/5) (approximately 1.2649)
For t ≥ 1: Use numerical methods to approximate the time(s) at which the particle is 4 cm from its starting position.