The velocity of a skateboard is v(t) = t^2 - 4 t + 3 m/s when moving in a straight line.
A. Find the the change in displacement of the skateboard between 4 seconds and 6 seconds. (Note this may or may not be negative, meaning it goes in the opposite direction, if so then be sure to include the sign.)
Change of displacement = m
B. Find the acceleration of the skateboard when t = 6 seconds.
Acceleration = m/s^2
A. Integrate the velocity
x = (t^3)/3-2t+3t
substitute 4 and 6
[(4^3)/3 - 2(4)+3(4)] - [(6^3)/3 - 2(6)+3(6)] = m?
B. get the derivatives of velocity
a = 2t^2 - 4
substitute t=6
2(6)^2-4 = m/s^2?
a = 2t-4, not 2t^2-4
To find the change in displacement of the skateboard between 4 seconds and 6 seconds, we need to integrate the velocity function with respect to time.
A. The velocity function is given as v(t) = t^2 - 4t + 3 m/s.
To find the change in displacement, we need to evaluate the integral of the velocity function from t = 4 to t = 6.
∫[4 to 6] v(t) dt = ∫[4 to 6] (t^2 - 4t + 3) dt
Integrating term by term, we get:
= [1/3 * t^3 - 2t^2 + 3t] evaluated from 4 to 6
= (1/3 * 6^3 - 2 * 6^2 + 3 * 6) - (1/3 * 4^3 - 2 * 4^2 + 3 * 4)
Simplifying further:
= (1/3 * 216 - 2 * 36 + 18) - (1/3 * 64 - 2 * 16 + 12)
= (72 - 72 + 18) - (64/3 - 32 + 12)
= 18 - (64/3 - 20)
= 18 - (64/3 - 60/3)
= 18 - 4/3
= 54/3 - 4/3
= 50/3
Therefore, the change in displacement of the skateboard between 4 seconds and 6 seconds is 50/3 m, or approximately 16.67 m.
B. To find the acceleration at t = 6 seconds, we need to differentiate the velocity function with respect to time.
The acceleration function is the derivative of the velocity function:
a(t) = d v(t) / dt = d / dt (t^2 - 4t + 3)
Differentiating term by term, we get:
a(t) = 2t - 4
Substituting t = 6 into the acceleration function:
a(6) = 2 * 6 - 4
a(6) = 12 - 4
a(6) = 8 m/s^2
Therefore, the acceleration of the skateboard at t = 6 seconds is 8 m/s^2.