A particle travels along a straight line with a velocity v=(12-3t^2) m/s, when t is in seconds. When t=1s, the particle is located 10m to the left of the origin. Determine the accelaration when t=4s, the displacement from t=0 to t=10s, and the distance the particle travels during this time period..
Thank you for the answer of my 1st question..hope you could help me w/ this one.tnx
v=12-3t^2
acceleration= -6t at t=4, acceleration is -24m/s^2
displacement=INTEGRAL velocity dt
= INT(12-3t^2)dt= 12t-t^3 from t=0 to 10
= 120-1000=-860
To find the acceleration when t = 4s, we need to differentiate the given velocity function v(t) with respect to time t.
Given: v(t) = 12 - 3t^2 m/s
Differentiating v(t) with respect to t, we get:
a(t) = d(v(t))/dt
= d(12 - 3t^2)/dt
= -6t
Now, substitute t = 4s into the acceleration function to find the acceleration at t = 4s:
a(4) = -6(4)
= -24 m/s^2
Therefore, when t = 4s, the acceleration is -24 m/s^2.
Now, let's find the displacement from t = 0s to t = 10s.
To find the displacement, we need to integrate the velocity function v(t) from t = 0s to t = 10s.
Given: v(t) = 12 - 3t^2 m/s
To find the displacement D, integrate v(t) with respect to t from t = 0s to t = 10s:
D = ∫(0s to 10s) (12 - 3t^2) dt
= ∫(0s to 10s) 12dt - ∫(0s to 10s) 3t^2 dt
= [12t] (0s to 10s) - [t^3] (0s to 10s)
= 12(10) - 3(10^3) - (12(0) - 3(0^3))
= 120 - 3000 - 0
= -2880 m
Therefore, the displacement from t = 0s to t = 10s is -2880 m (to the left).
To find the distance the particle travels during this time period, we can take the absolute value of the displacement.
Distance = |Displacement|
= |-2880|
= 2880 m
Therefore, the particle travels a distance of 2880 meters during the time period from t = 0s to t = 10s.
To determine the acceleration when t=4s, we need to find the derivative of the velocity with respect to time (t):
Given: v = 12 - 3t^2
To find the acceleration a, we take the derivative of v with respect to t:
a = d/dt (v) = d/dt (12 - 3t^2)
Differentiating the equation term by term, we get:
a = 0 - 6t
When t=4s, we can substitute the value into the equation to find the acceleration:
a = 0 - 6(4) = -24 m/s^2
Thus, the acceleration when t=4s is -24 m/s^2.
To determine the displacement from t=0 to t=10s, we need to calculate the integral of the velocity from t=0 to t=10:
Given: v = 12 - 3t^2
To find the displacement D, we integrate v with respect to t over the interval from t=0 to t=10:
D = ∫(12 - 3t^2) dt (evaluated from t=0 to t=10)
Integrating term by term, we get:
D = 12t - t^3
Substituting the limits of integration:
D = (12(10) - 10^3) - (12(0) - 0^3)
D = (120 - 1000) - (0 - 0)
D = -880 - 0
D = -880 m
Thus, the displacement from t=0 to t=10s is -880 m (to the left of the origin).
To determine the distance the particle travels during the time period from t=0 to t=10s, we need to find the integral of the absolute value of the velocity from t=0 to t=10:
Given: v = 12 - 3t^2
To find the distance traveled, we integrate the absolute value of v with respect to t over the interval from t=0 to t=10:
Distance = ∫|12 - 3t^2| dt (evaluated from t=0 to t=10)
We need to split this integral into two parts, corresponding to the different regions where the expression inside the absolute value function changes sign:
For t in [0, 2], v = 12 - 3t^2
For t in [2, 10], v = 3t^2 - 12
Integrating the first part, we get:
∫(12 - 3t^2) dt = 12t - t^3/3 + C1
Integrating the second part, we get:
∫(3t^2 - 12) dt = t^3 - 12t + C2
To find the distance, we need to convert any negative values obtained from evaluating the definite integrals to positive:
Distance = |12(10) - (10^3/3) + C1| + |(10^3 - 12(10) + C2)|
The constants C1 and C2 cancel out when evaluating the absolute value. For simplicity, let's assume C1 and C2 are both equal to zero.
Distance = |120 - (1000/3)| + |(1000 - 120)|
Distance = |360 - 1000/3| + |880|
Distance = |-640/3| + 880
Distance = 640/3 + 880
Distance = 2560/3 + 880
Distance ≈ 1426.67 + 880
Distance ≈ 2306.67 m
Thus, the distance the particle travels during the time period from t=0 to t=10s is approximately 2306.67 m.