A particle moves according to a law of motion s=f(t), t is less than or equal to

Question

A particle moves according to a law of motion s=f(t), t is less than or equal to 0, where t is measured in seconds and s in feet. f(t)=9t/(t^2+9).   a) Find the velocity at time t. b) What is the velocity after 1 second. c) when is the particle at rest? d)When is the particle moving in the positive direction? e) Find the total distance traveled during the first 6 seconds. f) Draw a diagram life Figure 2 to illustrate the motion of the particle. g)Find the acceleration at time t and after 1 second. h) Graph the position, velocity and acceleration functions for 0 is less than or equal to t which is less than or equal to 6. i) When is the particle speeding up? When is it slowing down?

Explanation / Answer

Here ,

s = 9t/(t^2 + 9)

a) velocity , v = ds/st

v = d/dt(9t/(t^2 + 9))

v = ( 9 *(t^2 + 9) - 2t * 9t) /(t^2 + 9)^2

v = 9 *(9 - t^2)/(t^2 + 9 )^2

b)

at t = 1 s

v = 9 *(9 - t^2)/(t^2 + 9 )^2

v = 9 *(9 - 1^2)/(1^2 + 9 )^2

v = 0.72 m/s

the velocity at t = 1 s is 0.72 m/s

c)

for the particle at rest

v = 0

9 *(9 - t^2)/(t^2 + 9 )^2 = 0

9 *(9 - t^2) = 0

sovling

t = 3 s

the velocity of particle is zero at t= 3 s

d)

for the particle to be moving in positive direction

v > 0

9 *(9 - t^2)/(t^2 + 9 )^2 > 0

9 *(9 - t^2) > 0

t < 3

the particle is moving in positive direction for t = 0 to t = 3 s

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