Question

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?

Answer 1

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

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

$x=3.77\ m$