A particle moves along straight line such that its displacement S meters from a given point is S = t^3 – 5t^2 + 4 whee t is time in seconds.
Find
(a) The displacement of particle at t = 5
(b) The velocity of the particle when t = 5
(c) The values of t when the particle is momentarily at rest
(d) The acceleration of the particle when t = 2
S(5) = 125 -125 + 4 = 4
v = 3 t^2 - 10 t
v(5) = 75 - 50
when is v = 0?
3 t^2 - 10 t = 0 = t(3 t - 10)
t = 0 or t - 10/3
a = 6 t - 10
a(2) = 12 - 10 = 2
To find the displacement, velocity, values of t when the particle is momentarily at rest, and acceleration, we need to differentiate the equation for displacement with respect to time.
The given equation for displacement is S = t^3 – 5t^2 + 4.
(a) To find the displacement of the particle at t = 5, substitute t = 5 into the equation:
S = (5)^3 – 5(5)^2 + 4
S = 125 – 125 + 4
S = 4
Therefore, the displacement of the particle at t = 5 is 4 meters.
(b) To find the velocity at t = 5, differentiate the equation for displacement with respect to t:
v = dS/dt
v = d/dt (t^3 – 5t^2 + 4)
v = 3t^2 – 10t
Substitute t = 5 into the derived equation:
v = 3(5)^2 – 10(5)
v = 75 – 50
v = 25
Therefore, the velocity of the particle when t = 5 is 25 meters per second.
(c) To find the values of t when the particle is momentarily at rest, we need to find when the velocity is zero.
Set the derived equation for velocity to zero:
0 = 3t^2 – 10t
Factor out t from the equation:
0 = t(3t – 10)
Set each factor equal to zero and solve for t:
t = 0 or 3t – 10 = 0
t = 0 or 3t = 10
t = 0 or t = 10/3
Therefore, the particle is momentarily at rest when t = 0 or t = 10/3 seconds.
(d) To find the acceleration at t = 2, differentiate the equation for velocity with respect to t:
a = dv/dt
a = d/dt (3t^2 – 10t)
a = 6t – 10
Substitute t = 2 into the derived equation:
a = 6(2) – 10
a = 12 – 10
a = 2
Therefore, the acceleration of the particle when t = 2 is 2 meters per second squared.
To find the answers to the given questions, we need to understand the concept of displacement, velocity, and acceleration.
(a) Displacement at time t = 5:
The displacement S can be found by substituting t = 5 into the equation S = t^3 - 5t^2 + 4.
So, S = 5^3 - 5(5^2) + 4
Simplifying further, S = 125 - 125 + 4
S = 4 meters.
(b) Velocity at time t = 5:
Velocity is the rate of change of displacement with respect to time. To find the velocity at a specific time, we need to find the derivative of the displacement equation with respect to time.
Taking the derivative of S = t^3 - 5t^2 + 4, we get:
V = dS/dt = 3t^2 - 10t
Substituting t = 5 into the velocity equation, we get:
V = 3(5^2) - 10(5)
V = 3(25) - 50
V = 75 - 50
V = 25 meters per second.
(c) Values of t when the particle is momentarily at rest:
When the particle is at rest, its velocity is zero. So, we can set the velocity equation equal to zero and solve for t.
3t^2 - 10t = 0
t(3t - 10) = 0
From this equation, we can have two possibilities:
t = 0 (at the beginning)
3t - 10 = 0
3t = 10
t = 10/3 (approximately 3.33 seconds)
Therefore, when t = 0 (at the start) and t = 10/3 (approximately 3.33 seconds), the particle is momentarily at rest.
(d) Acceleration at t = 2:
Acceleration is the rate of change of velocity with respect to time. To find the acceleration at a specific time, we need to find the derivative of the velocity equation with respect to time.
Taking the derivative of V = 3t^2 - 10t, we get:
A = dV/dt = 6t - 10
Substituting t = 2 into the acceleration equation, we get:
A = 6(2) - 10
A = 12 - 10
A = 2 meters per second squared.