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Effectus [21]
3 years ago
8

An object’s velocity is measured to be vx(t) = α - βt2, where α = 4.00 m/s and β = 2.00 m/s3. At t = 0 the object is at x = 0. (

a) Calculate the object’s position and acceleration as functions of time. (b) What is the object’s maximum positive displacement from the origin?
Physics
1 answer:
Leona [35]3 years ago
7 0

Answer:

Explanation:

Given

v_x(t)=\alpha -\beta t^2

\alpha =4\ m/s

\beta =2\ m/s^3

v_x(t)=4-2t^2

v=\frac{\mathrm{d} x}{\mathrm{d} t}

\int dx=\int \left ( 4-2t^2\right )dt

x=4t-\frac{2}{3}t^3

acceleration of object

a=\frac{\mathrm{d} v}{\mathrm{d} t}

a=-4t

(b)For maximum positive displacement velocity must be zero at that instant

i.e.v=0

4-2t^2=0

t=\pm \sqrt{2}

substitute the value of t

x=4\times \sqrt{2}-\frac{2}{3}\times 2\sqrt{2}

x=3.77\ m

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