A particle travels along a straight line with a velocity v=(12-3t^2)m/s.where t is in the seconds. When t=1s.the particle is located 10m to the left of the origin. Determine the acceleration when t=4s.the displacement from t=0s to t=10s.and the distance the particle travels during this time period.
the distance s(t) = 12t - t^2 + c
for some c.
Since s(1) = -10, c=-21, and so
s(t) = -21 + 12t - t^2
a = dv/dt = -6t
so just plug in t=4
plug in t= 0 and 10 to find where the particle is at each time, then get the displacement by evaluating s(10)-s(0)
To determine the acceleration when t=4s, we need to find the derivative of the velocity function v(t) with respect to time t.
Given that v(t) = 12 - 3t^2, we calculate the derivative as follows:
a(t) = d/dt (v(t))
= d/dt (12 - 3t^2)
= -6t
Now we can substitute t = 4s into the acceleration equation:
a(t=4s) = -6(4)
= -24 m/s^2
Therefore, the acceleration when t=4s is -24 m/s^2.
To find the displacement from t=0s to t=10s, we need to integrate the velocity function v(t) with respect to time t over the interval [0, 10].
Given that v(t) = 12 - 3t^2, we calculate the integral as follows:
s(t) = ∫[0, 10] (v(t)) dt
= ∫[0, 10] (12 - 3t^2) dt
= [12t - t^3]_[0, 10]
= (12(10) - 10^3) - (12(0) - 0^3)
= 120 - 1000 - 0
= -880 m
Therefore, the displacement from t=0s to t=10s is -880 m.
To find the distance the particle travels during this time period, we need to calculate the definite integral of the magnitude of the velocity function v(t) with respect to time t over the interval [0, 10].
Given that v(t) = 12 - 3t^2, we calculate the integral as follows:
d(t) = ∫[0, 10] |v(t)| dt
To evaluate this integral, we need to split it into two parts based on the sign of the velocity function over the given interval. Since the velocity v(t) is positive for t ≤ 2 and negative for t > 2, we have
d(t) = ∫[0, 2] (v(t)) dt + ∫[2, 10] (-v(t)) dt
Using the given velocity function v(t) = 12 - 3t^2, we calculate the integrals as follows:
∫[0, 2] (v(t)) dt = ∫[0, 2] (12 - 3t^2) dt
= [12t - t^3/3]_[0, 2]
= (12(2) - (2)^3/3) - (12(0) - (0)^3/3)
= 12(2) - 8/3
= 24 - 8/3
= 72/3 - 8/3
= 64/3
∫[2, 10] (-v(t)) dt = - ∫[2, 10] (12 - 3t^2) dt
= -∫[2, 10] (12 - 3t^2) dt
= -[12t - t^3/3]_[2, 10]
= -(12(10) - (10)^3/3) - (12(2) - (2)^3/3)
= -120 + 1000/3 - 24 + 8/3
= -360/3 + 1000/3 - 72/3 + 8/3
= 568/3
Therefore, the distance the particle travels during the time period from t=0s to t=10s is:
d(t=0s to t=10s) = |∫[0, 2] (v(t)) dt + ∫[2, 10] (-v(t)) dt|
= |64/3 + 568/3|
= |632/3|
= 632/3 m
Therefore, the distance the particle travels during this time period is 632/3 meters.
To find the acceleration when t = 4s, we need to find the derivative of the velocity function with respect to time.
Given velocity function:
v = 12 - 3t^2
Differentiating v with respect to t, we get:
a = dv/dt = d/dt(12 - 3t^2)
= 0 - (3 * 2t)
= -6t
Substituting t = 4s, we find the acceleration:
a = -6 * 4
a = -24 m/s^2
Therefore, the acceleration when t = 4s is -24 m/s^2.
To find the displacement from t = 0s to t = 10s, we need to integrate the velocity function with respect to time.
Given velocity function:
v = 12 - 3t^2
Integrating v with respect to t, we get:
s = ∫(12 - 3t^2) dt
= 12t - t^3 + C
To find the constant of integration (C), we'll use the information given:
When t = 1s, s = 10m
Substituting these values into the equation, we have:
10 = 12(1) - (1)^3 + C
10 = 12 - 1 + C
C = -1
So, the displacement from t = 0s to t = 10s is:
s = 12t - t^3 - 1
To find the distance the particle travels during this time period (from t = 0s to t = 10s), we need to consider the absolute value of the displacement.
Taking the absolute value, we get:
distance = |12t - t^3 - 1|
Since t ranges from 0s to 10s, we'll evaluate this expression for those values:
distance = |12(0) - (0)^3 - 1| + |12(10) - (10)^3 - 1|
= |-1| + |120 - 1000 - 1|
= 1 + |-881|
= 1 + 881
= 882 meters
Therefore, the distance the particle travels during this time period is 882 meters.