Suppose that a particle moves according to the law of motion s\' = ((t^2 + 9)*4-

Question

Suppose that a particle moves according to the law of motion

s' = ((t^2 + 9)*4-(4t)(2t)) / ((t^2 + 9)^2) , t ? 0



PART A: Use interval notation to indicate when the particle is moving in the positive direction.

PART B: Find the total distance traveled during the first 8 seconds
Suppose that a particle moves according to the law of motion

s' = ((t^2 + 9)*4-(4t)(2t)) / ((t^2 + 9)^2) , t ? 0



PART A: Use interval notation to indicate when the particle is moving in the positive direction.

PART B: Find the total distance traveled during the first 8 seconds

Explanation / Answer

part a

expanding it we get

=(4t^2 + 36 - 8t^2) / ( t^4 + 81 + 18t^2)

= (36-4t^2) / ((t^2+9)^2)

= ((6+2t)(6-2t)) / ((t^2+9)^2)

dierentiating it we get the velocity

(-8t( (t^2+9)^2) - 2(t^2+9)2t ) / ((t^2 +9)^2)

expanding and cancelling t^2 +9

-8t^3-76t / t^2 + 9

the velocity is becoing zer0 in

8t^2+76 =0 t=0

so velocity is only changing when t = 0

and at all other time it is negetive so the particle will keep moving in the negative direction

part b

at t=0

s= 36/81

till t= 3 sec

s= 0

at t=8

-220/5329

so total distance travelled is

36/81 + 220/5329

0.32 + 0.041

0.361 units

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