a particle moves along the x-axis in such a way that its acceleration at any time t is given by a(t)= 6t - 18. At time t = 0, the velocity v(t) = 24. and at time t = 1, the position x(t) = 20
i have to find the expression for the velocity and what values t is at rest
since a(t) = 6t -18
v(t) = 3t^2 - 18 + c
but we are told v(0) = 24
so 24 = 0 + 0 + c, ----> c = 24
if v(t) = 3t^2 - 18t + 24
x(t) = t^3 - 9t^2 + 24t + k
but we are told x(1) = 20
so 20 = 1 - 9 + 24 + k , ---> k = 4
x(t) = t^3 - 9t^2 + 24t + 4
Your last sentence is confusing.
Did you mean " ... and what values of t the particle is at rest" ?
anyway, since you now have all 3 equations, I am pretty sure you can proceed from there
To find the expression for velocity, we need to integrate the acceleration function with respect to time.
Integrating a(t) = 6t - 18 with respect to t gives us:
v(t) = 3t^2 - 18t + C
We know that at time t = 0, the velocity v(t) is equal to 24. Substituting these values into the expression for v(t):
24 = 3(0)^2 - 18(0) + C
24 = C
Therefore, the expression for velocity is:
v(t) = 3t^2 - 18t + 24
To find the values of t when the particle is at rest, we need to find when the velocity v(t) is equal to zero.
Setting v(t) = 0:
0 = 3t^2 - 18t + 24
To solve this quadratic equation, we can factor it:
0 = 3(t^2 - 6t + 8)
0 = 3(t - 4)(t - 2)
So, t = 4 or t = 2.
Therefore, the particle is at rest at times t = 4 and t = 2.
To find the expression for velocity, we integrate the given acceleration function, a(t), with respect to time t.
∫ a(t) dt = ∫(6t - 18) dt
To find the expression for velocity, we integrate each term separately:
∫ 6t dt = 3t^2 + C1
∫ -18 dt = -18t + C2
where C1 and C2 are integration constants.
Combining the two integrals:
v(t) = 3t^2 - 18t + C
Given that the velocity v(t) = 24 at t = 0, we can substitute these values into the expression for velocity:
24 = 3(0)^2 - 18(0) + C
24 = 0 + 0 + C
C = 24
Therefore, the expression for velocity, v(t), is:
v(t) = 3t^2 - 18t + 24
To find when the particle is at rest, we need to determine the values of t at which the velocity, v(t), is equal to zero.
Setting v(t) = 0:
3t^2 - 18t + 24 = 0
We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Factoring the equation, we get:
3(t^2 - 6t + 8) = 0
Factoring further, we have:
3(t - 2)(t - 4) = 0
This equation will be true when either:
t - 2 = 0 or t - 4 = 0
Solving for t in each case, we find:
t = 2 or t = 4
Therefore, the particle is at rest at t = 2 and t = 4.