Answer to Question #167028 in Calculus for Angelo

Question #167028

If π‘Ž = 4𝑑^βˆ’3/2 , 𝑠 = 16 π‘€β„Žπ‘’π‘› 𝑑 = 4, π‘Žπ‘›π‘‘ 𝑠 = 25 π‘€β„Žπ‘’π‘› 𝑑 = 6, find the equation of motion 𝑠 = 𝑓(𝑑) and the velocity function 𝑣(𝑑).


1
Expert's answer
2021-03-01T07:15:40-0500

Acceleration aa is given as


a=dvdtβ€…β€ŠβŸΉβ€…β€Šdv=aβ€…β€Šdtβ€…β€ŠβŸΉβ€…β€Šv=∫aβ€…β€Šdta = \dfrac{dv}{dt} \implies dv = a\;dt \implies v = \int a\;dt

The function of the velocity (vv) is given as:


v=∫4tβˆ’3/2dt=4∫tβˆ’3/2dt=4∫1t32dt=4(βˆ’2t)v=βˆ’8t+cv =\int 4t^{-3/2} dt\\ = 4 \int t^{-3/2} dt\\ = 4\int \dfrac{1}{t^\frac{3}{2}} dt\\ = 4 \Bigg(-\dfrac{2}{\sqrt{t}}\Bigg)\\ v=-\dfrac{8}{\sqrt{t}} + c

Velocity (vv) is given as:


v=dsdtβ€…β€ŠβŸΉβ€…β€Šds=vβ€…β€Šdtβ€…β€ŠβŸΉβ€…β€Šs=∫vβ€…β€Šdtv = \dfrac{ds}{dt} \implies ds = v\; dt \implies s = \int v\;dt

The function of the equation of motion can be obtained as follows:


s=∫(βˆ’8t+c)dt=βˆ’8(2t)+cts=βˆ’16t+ct+c1s= \int \Big(-\dfrac{8}{\sqrt{t}} + c \Big) dt\\ = -8 (2\sqrt{t}) + ct\\ s = - 16\sqrt{t} + ct +c_1

When s = 16 and t = 4:


16=βˆ’16(4)+c(4)+c1=βˆ’16(2)+4c+c116=βˆ’32+4c+c1β€…β€ŠβŸΉβ€…β€Š4c+c1=48…(eqnβ€…β€Š1)16 = -16(\sqrt{4})+c(4) + c_1\\ = -16(2) + 4c + c_1\\ 16= -32 +4c +c_1\\ \implies 4c+c_1 = 48 \qquad\dots (eqn\;1)

When s = 25 and t = 6:


25=βˆ’16(6)+c(6)+c1=βˆ’16(6)+6c+c125=βˆ’16(2.45)+6c+c16c+c1=25+39.2β€…β€ŠβŸΉβ€…β€Š6c+c1=64.2…(eqnβ€…β€Š2)25 = -16(\sqrt{6})+c(6) + c_1\\ = -16(\sqrt{6}) + 6c + c_1\\ 25= -16(2.45) +6c +c_1\\ 6c+c_1 = 25+39.2 \\ \implies 6c+c_1 = 64.2 \qquad\dots (eqn\;2)

Solving eqn 1 and eqn 2 simultaneously:


4c+c1=48(βˆ’)6c+c1=64.2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’2c=βˆ’16.2c=8.1\quad4c+c_1=48\\ (-) 6c+c1=64.2\\ ---------\\ -2c=-16.2\\ c= 8.1

Hence we find c1c_1 :


4(8.1)+c1=4832.4+c1=48c1=48βˆ’32.4c1=15.64(8.1) +c_1 = 48\\ 32.4 +c_1 = 48\\ c_1 = 48 -32.4\\ c_1=15.6

Therefore, the equation of the motion s=f(t)s = f(t) is:


s=f(t)=βˆ’16t+8.1t+15.6s=f(t)=-16\sqrt{t}+8.1t+15.6

The velocity v(t)v(t) is:


v(t)=βˆ’8t+8.1v(t) = -\dfrac{8}{\sqrt{t}} + 8.1


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
16.03.21, 18:13

Dear Meep Melphy, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

Meep Melphy
16.03.21, 11:00

Thank you.

Leave a comment

LATEST TUTORIALS
New on Blog