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elena55 [62]
3 years ago
14

The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve

locity of 4 ft/s , which is increasing at the rate of v˙=(0.004t)ft/s2, where t is in seconds.A-Determine the magnitude of the velocity when it has traveled three-fourths the way around the track.B-Determine the magnitude of the acceleration when it has traveled three-fourths the way around the track.
Engineering
1 answer:
stellarik [79]3 years ago
8 0

Answer:

Explanation:

Given

velocity at A is v=4\ ft/s

For r=500\ ft

velocity is increasing at \dot{v}=0.004t\ ft/s^2

Tangential acceleration is given by

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

a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}

\int 0.004tdt=\int dv

\int dv=\int 0.004tdt

v=0.002t^2+c

at t=0\ v=4\ ft/s

4=0.002\cdot 0+c

c=4\ ft/s

thus v=0.002t^2+4

Velocity in terms of Displacement is given by

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

\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt

\Rightarrow s=\frac{0.002t^3}{3}+4t

When car has traveled \frac{3}{4} th of distance i.e.

s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}

s=750\pi

750\pi =\frac{0.002t^3}{3}+4t

\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0

on solving we get t=139.23\ s

Thus velocity at t=139.23\ s

v=42.76\ s

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}

a_n=3.658\ m/s^2

Tangential acceleration a_t at t=139.23\ s

a_t=0.556\ m/s^2

Net acceleration a_t=\sqrt{(a_n)^2+(a_t)^2}

a_n=\sqrt{(3.658)^2+(0.556)^2}

a_n=3.7\ m/s^2

   

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