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?
a = 24-15√t
v = 24t-10t^(3/2)
s = 12t^2 - 4t^(5/2)
s=0 when 12-4√t = 0, or t=9
a(9) = 24-15√9 = -21
To find the acceleration at which the particle returns to the starting point, we need to determine the time at which the particle reaches the starting point.
The particle reaches the starting point when the distance traveled is zero. Therefore, we can set up the equation:
Distance = 0
To find the time at which the particle reaches zero distance, we can integrate the given acceleration function:
∫ (24 - 15√t) dt
This integration will give us the velocity function, which represents how the velocity of the particle changes over time.
After integrating the given acceleration function, we get:
∫ (24 - 15√t) dt = 24t - 10t^(3/2) + C
Where C is the constant of integration.
The particle starts from rest, meaning its initial velocity is zero. Therefore, we can set up the equation:
Velocity = 0
24t - 10t^(3/2) + C = 0
To solve for t, we need to rearrange the equation:
10t^(3/2) - 24t + C = 0
Since the particle is at the starting point, the time t must be positive. To find the value of C, we substitute t = 0 into the equation:
10(0)^(3/2) - 24(0) + C = 0
0 + C = 0
C = 0
Now we have the equation:
10t^(3/2) - 24t = 0
Factoring out t:
t(10t^(1/2) - 24) = 0
From this equation, we can see that t = 0 or 10t^(1/2) - 24 = 0
Solving for t in the second equation:
10t^(1/2) = 24
t^(1/2) = 24/10
t^(1/2) = 2.4
Squaring both sides:
t = (2.4)^2
t = 5.76 seconds
Therefore, the particle returns to the starting point after 5.76 seconds.
To find the acceleration at this time, we can substitute t = 5.76 back into the given acceleration function:
Acceleration = 24 - 15√(5.76)
Acceleration ≈ 24 - 15(2.4)
Acceleration ≈ 24 - 36
Acceleration ≈ -12 feet/second^2
Thus, the particle returns to the starting point with an acceleration of approximately -12 feet/second^2.