# applications of differentiation-velocity+acceleration?

a body is projected through a resistance medium and travels a distance x metres in a time t seconds according to the equation:

x= 36t-3/8 t^2

(i)how long does it take for the body to come to rest?

(ii)how far does the body travel before it comes to rest?

(iii)what are the velocity and acceleration when time is 16 seconds?

would appreciate the steps on how to solve this :)

Update:

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Relevance
• Ari
Lv 6

i) when something is at rest, it's velocity is 0

we need to find velocity

v= x ' = 36 - (3/8)*2 t

v= 36 - (3/4)t

set it equal to 0

36 - (3/4)t = 0

(3/4)t = 36

t = 48

it takes 48 seconds for the body to come back to rest.

ii) since the body doesn't change direction from t=0 to t=48, the distance traveled is just :

x(48) - x(0) = 1710 - 0 = 864.

iii) v (16) = 36 - (3/4)*16 = 36-12 = 24 m/s

we need to find acceleration.

a = v ' = -3/4

a= -3/4 irrespective of the value of t, so a(16) is certainly also -3/4.

The position function, x(t), is given by

x(t) = 36t - 3/8 t², or

x(t) = - 3/8 t² + 36t

The veolcity function is given by the derivative, x'(t), of the position function:

x'(t) = - 3/4 t + 36

The body is at rest when the velocity = 0, so

x'(t) = 0, or

- 3/4 t + 36 = 0

- 3/4 t = - 36

t = - 36 / (- 3/4)

t = - 36(- 4/3)

t = 48

It takes 48 secs. for the body to come to rest.

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The distance, x(t), at time t is given by

x(t) = - 3/8 t² + 36t

x(t) = - 3/8(48)² + 36(48)

x(t) = - 3/8(2304) + 1728

x(t) = - 864 + 1728

x(t) = 864

The body travels 864 meters before it comes to rest.

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The velocity, x'(t), at 16 secs. is given by

x'(t) = - 3/4(16) + 36

x'(t) = - 12 + 36

x(t)' = 24

The velocity after 16 secs. = 24 m./sec.

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The acceleration, x''(t), is given by the derivative of the velocity function:

x''(t) = - 3/4

The acceleration = - 0.75 m./sec.² (Negative because the body is decelerating).

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Source(s): 8/8/10
• TomV
Lv 7

v = dx/dt = 36 - 3t/4

a = dv/dt = -3/4

i) v = 0 = 36-3t/4

3t/4 = 36

t = 48 seconds

ii) x(48) = 36(48) - (3/8)(48²) = 864 meters

iii) acceleration is constant at -3/4 m/t²

v(16) = 36-3(16/4) = 24 m/s

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