A particle is moving along a straight line and its position is given by the relation x=( t3-6t2-15t+40)mm. Find:-
(a). The time at which velocity is zero.
(b). Position & displacement of the particle at that point.
(c). Acceleration for the particle at that line.
Given a position function x(t), then
x'(t)=dx(t)/dt=velocity, and
x"(t)=d²x(t)/dt²=acceleration.
where x' and x" are the first and second derivatives with respect to time.
(a) solve for x'(t0)=0
(b) evaluate position = x(t0), and displacement = x(t0)-x(0).
(c) evaluate x"(t0)
To find the time at which the velocity is zero, we need to determine the velocity function first. Velocity is the derivative of position with respect to time.
Given: x = t^3 - 6t^2 - 15t + 40
Step 1: Take the derivative of x with respect to t to obtain the velocity function.
v = dx/dt = d/dt (t^3 - 6t^2 - 15t + 40)
Step 2: Differentiate each term of the equation.
v = 3t^2 - 12t - 15
(a) To find the time at which velocity is zero, set the velocity function equal to zero and solve for t.
0 = 3t^2 - 12t - 15
This is a quadratic equation, so let's solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = -12, and c = -15.
t = (-(-12) ± √((-12)^2 - 4 * 3 * (-15))) / (2 * 3)
t = (12 ± √(144 + 180)) / 6
t = (12 ± √324) / 6
t = (12 ± 18) / 6
Simplifying further:
t1 = (12 + 18) / 6 = 30 / 6 = 5
t2 = (12 - 18) / 6 = -6 / 6 = -1
Therefore, the time at which velocity is zero is t = 5.
(b) To find the position and displacement of the particle at that point, substitute t = 5 into the position function:
x = t^3 - 6t^2 - 15t + 40
x = 5^3 - 6(5)^2 - 15(5) + 40
x = 125 - 150 - 75 + 40
x = -60
The position of the particle at that time is -60 mm.
Displacement is the change in position. To find displacement, calculate the position of the particle at t=0:
x = (0^3) - 6(0^2) - 15(0) + 40
x = 0 - 0 - 0 + 40
x = 40
Therefore, the displacement of the particle at that time is given by the difference in positions:
Displacement = -60 - 40 = -100 mm
(c) Acceleration is the derivative of velocity with respect to time. To find acceleration at that time, take the derivative of the velocity function:
a = dv/dt = d/dt (3t^2 - 12t - 15)
Differentiating each term:
a = 6t - 12
Substitute t = 5 into the acceleration function:
a = 6(5) - 12
a = 30 - 12
a = 18
Therefore, the acceleration of the particle at that time is 18 mm/s^2.
To find the time at which velocity is zero, we need to find the derivative of the position function and set it equal to zero.
(a) Velocity of the particle can be found by taking the derivative of the position function with respect to time (t):
v(t) = x'(t) = 3t^2 - 12t - 15
To find the time at which velocity is zero, we set v(t) = 0 and solve for t:
3t^2 - 12t - 15 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula.
Factoring the equation:
3(t - 5)(t + 1) = 0
Setting each factor equal to zero:
t - 5 = 0 or t + 1 = 0
Solving for t in each case:
t = 5 or t = -1
Therefore, the particle's velocity is zero at t = 5 mm and t = -1 mm.
(b) To find the position and displacement of the particle at t = 5 mm, we substitute t = 5 into the position function:
x(5) = (5^3) - 6(5^2) - 15(5) + 40
Evaluating the expression:
x(5) = 125 - 150 - 75 + 40
x(5) = -60 mm
The position of the particle at t = 5 mm is -60 mm.
Displacement can be calculated by finding the difference between the initial and final positions:
Displacement = x(final) - x(initial)
= x(5) - x(0)
= -60 - 40
= -100 mm
Therefore, the displacement of the particle at t = 5 mm is -100 mm.
(c) Acceleration of the particle can be found by taking the derivative of the velocity function:
a(t) = v'(t) = 6t - 12
To find the acceleration at t = 5 mm, substitute t = 5 into the acceleration function:
a(5) = 6(5) - 12
a(5) = 30 - 12
a(5) = 18 mm/s^2
Therefore, the acceleration of the particle at t = 5 mm is 18 mm/s^2.