Respuesta :
Answer:
a)36.33 m
b)346.7 m/s
c)7 s
d)128.33 m/s
Explanation:
Given:
Variation of acceleration with time, [tex] a(t)= 10 (-t + 2 ) (t-5)+ 100[/tex]
Now using the calculus we have
[tex]a(t)=\dfrac{dv}{dt}=10(-t+2)(t-5)+100\\\\\int\limits^v_0 {} \, dv =\int\limits^t_0 {(10(-t+2)(t-5)+100)} \, dt\\v=-[tex]\dfrac{dx}{dt}=-10\left (\dfrac{t^3}{3}-\dfrac{7t^2}{2}+10t \right )[/tex]
Again using calculus we have
[tex]v=-\dfrac{dx}{dt}=-10\left (\dfrac{t^3}{3}-\dfrac{7t^2}{2}+10t \right ) +100t\\\\\int\limits^x_3 {} \, dx =\int\limits^t_0 {-10\left (\dfrac{t^3}{3}-\dfrac{7t^2}{2}+10t \right ) +100t} \, dt\\ x=-10\left (\dfrac{t^4}{4}-\dfrac{7t^3}{6}+5t^2 \right )+50t^2+3[/tex]
a) At t =2 we have displacement
[tex] x=-10\left (\dfrac{2^4}{4}-\dfrac{7\times2^3}{6}+5\rimes2^2 \right )+50\times2^2+3\\\\x=36.33\ \rm m[/tex]
b) Now Let v be the velocity at t=4 s then
[tex]v=-10\left (\dfrac{t^3}{3}-\dfrac{7t^2}{2}+10t \right )\\\\v=-10\left (\dfrac{4^3}{3}-\dfrac{7\times 4^2}{2}+10\times 4\right )\\\\v=346.67\ \rm m/s [/tex]
c Let t be the time at which the Maximum displacement occurs
Then to Find the time maximum value we will equate the derivative of x with time to 0
[tex]\dfrac{dx}{dt}=-10\left (\dfrac{t^3}{3}-\dfrac{7t^2}{2}+10t \right ) +100t=0\\t=0\ or\ 7[/tex]
At t=0 we have minimum value of the displacement as the body starts at this time. So the particle will have maximum displacement at t=7 s.
d) In order to find the maximum value of velocity we will equate the derivative of v with respect to time equal to 0
[tex]=\dfrac{dv}{dt}=10(-t+2)(t-5)+100=0\\\\t=7\ \rm s\\\\v_{max}=-10\left (\dfrac{t^3}{3}-\dfrac{7t^2}{2}+10t \right )\\\\v=-10\left (\dfrac{7^3}{3}-\dfrac{7\times7^2}{2}+10\times 7 \right )+100\times 7=128.33\ \rm m/s[/tex]
