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?
help please thank you :-)
d=0=1/2 a t^2=12t^2-7.5t^3/2
t^2(12-7.5sqrtt)=0
sqrtt=12/7.5
t= that squared.
check my thinking
To determine the acceleration at which the particle returns to the starting point, we need to find the time at which the particle reaches the starting point and then substitute that time into the given acceleration equation to find the corresponding acceleration.
First, let's find the time when the particle returns to the starting point. When the particle returns to the starting point, the displacement is zero. So, we can set the expression for the displacement equal to zero and solve for the time.
The expression for displacement can be found by integrating the given acceleration equation with respect to time:
∫(24-15√t) dt = 0
To find the integral of the acceleration equation, we can first break it into its individual terms:
∫24 dt - ∫15√t dt = 0
Integrating both terms:
24t - (2/3)(15/2)t^(3/2) = 0
24t - 15t^(3/2) = 0
Now let's solve for t:
15t^(3/2) = 24t
Dividing both sides by t:
15t^(1/2) = 24
Now, square both sides:
15t = 576
t = 576/15
t = 38.4 seconds
So, the particle will return to the starting point after 38.4 seconds.
Now, substitute this value of time back into the given acceleration equation to find the acceleration at this time:
Acceleration at t = 38.4 seconds:
a = 24 - 15√(38.4)
Now, simply calculate this expression to find the acceleration at the time when the particle returns to the starting point.