A particle is moving along a straight line and its position is given by the relation
x=(t3- 6t2- 15t+40) m
FIND a) The time at which velocity is Zero,
b) Position and displacement of the particle at that point.
c) Acceleration for the particle at that line.
To find the time at which the velocity is zero, we need to find when the derivative of the position function is zero.
a) First, we need to differentiate the position function with respect to time (t) to get the velocity function:
v(t) = d/dt(x) = d/dt(t^3 - 6t^2 - 15t + 40)
Differentiating each term separately:
v(t) = 3t^2 - 12t - 15
To find the time at which velocity is zero, we set v(t) equal to zero and solve for t:
0 = 3t^2 - 12t - 15
This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Let's use 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
This gives us two possible solutions for t:
t1 = (12 + 18) / 6
t1 = 30 / 6
t1 = 5
t2 = (12 - 18) / 6
t2 = -6 / 6
t2 = -1
Since time cannot be negative in this case, we discard t2 and conclude that the time at which velocity is zero is t = 5 seconds.
b) To find the position of the particle at that point, we substitute t = 5 into the position function:
x(5) = (5^3 - 6(5)^2 - 15(5) + 40) m
x(5) = (125 - 150 - 75 + 40) m
x(5) = (-60) m
Therefore, the position of the particle at the point where the velocity is zero is x = -60 meters.
To find the displacement, we calculate the difference between the final position and the initial position:
Displacement = x(final) - x(initial)
Assuming the initial position is x(0) = (0^3 - 6(0)^2 - 15(0) + 40) m = 40 m,
Displacement = (-60) - 40
Displacement = -100 meters
So, the displacement of the particle at the point where the velocity is zero is -100 meters.
c) To find the acceleration at that point, we differentiate the velocity function with respect to time:
a(t) = d/dt(v) = d/dt(3t^2 - 12t - 15)
Differentiating each term separately:
a(t) = 6t - 12
To find the acceleration at t = 5, we substitute t = 5 into the acceleration function:
a(5) = 6(5) - 12
a(5) = 30 - 12
a(5) = 18 m/s^2
Therefore, the acceleration for the particle at the point where the velocity is zero is 18 m/s^2.