derivitive of velocity is displacement(distance travelled)
dy/dx = 12t^3-6t
plug in 4, and you get 744 units
You need to integrate the velocity function, which will give you the distance function, and then use the notion of the definite integral:
v(t) = 3t^4 - 3t^2
â«v(t) dt = â«(3t^4 - 3t^2) dt
= (3/5)t^5 - (3/3)t^3
Evaluating from t = 0 to t = 4:
[(3/5)(4)^5 - 4^3)] - [(3/5)*0^5 - 0^3]
= (3/5)(1024) - 64
= (3072/5) - (320/5)
= 2752/5 = 550.4
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derivitive of velocity is displacement(distance travelled)
dy/dx = 12t^3-6t
plug in 4, and you get 744 units
You need to integrate the velocity function, which will give you the distance function, and then use the notion of the definite integral:
v(t) = 3t^4 - 3t^2
â«v(t) dt = â«(3t^4 - 3t^2) dt
= (3/5)t^5 - (3/3)t^3
Evaluating from t = 0 to t = 4:
[(3/5)(4)^5 - 4^3)] - [(3/5)*0^5 - 0^3]
= (3/5)(1024) - 64
= (3072/5) - (320/5)
= 2752/5 = 550.4