Question

A particle is moving along a straight line with an initial velocity of 6 m/s when it is subjected to a deceleration of a- (-1.5v12) m/s2, where v is in m/s. Determine how far it travels before it stops. How much time does this take?

Answer 1

Answer:

s= 6.53 m

t=3.27 s

Explanation:

velocity = 6 m/s

deceleration = -1.5$v^\frac{1}{2}$

$a=-1.5v^\frac{1}{2}\\v\frac{\mathrm{d} v}{\mathrm{d} s}=-1.5v^\frac{1}{2}\\-1.5ds=v^\frac{1}{2}dv\\\int\ {-1.5} \, ds= \int\ v^\frac{1}{2}dv\\-1.5s=\frac{2}{3}\times v^{\frac{3}{2} }$

now inserting value of v=6s we get distance(s)

s= 6.53 m ( distance cannot be negative)

now for time calculation we know that

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

$-1.5v^\frac{1}{2} =\frac{\mathrm{d}v }{\mathrm{d} t}\\-1.5dt=\frac{dv}{v^\frac{1}{2}} \\\int -1.5 dt=\int v^{-\frac{1}{2}}dt \\1.5t=2v^\frac{1}{2}\\t=\frac{4}{3}v^\frac{1}{2}$

putting value of v=6s

t=3.27 s (time cannot be negative)