Question

Given a particle that follows the acceleration a(t) 10 (-t + 2 ) (t-5)+ 100 m/s2, find: a. Find the displacement at 2 seconds. Assume from rest and a starting point of 3m. b. Find the velocity at 4 seconds. Assume an initially at rest. c. Find the time at which maximum displacement occurs (use calculus because it is way easier, not kinematics). d. Find the value of maximum velocity over the interval 0

Answer 1

a)36.33 m

b)346.7 m/s

c)7 s

d)128.33 m/s

Step-by-step explanation:

Given:

Variation of acceleration with time,
`a(t)= 10 (-t + 2 ) (t-5)+ 100`

Now using the calculus we have

`a(t)=(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](dx)/(dt)=-10\left ((t^3)/(3)-(7t^2)/(2)+10t \right )`

Again using calculus we have

`v=-(dx)/(dt)=-10\left ((t^3)/(3)-(7t^2)/(2)+10t \right ) +100t\\\\\int\limits^x_3 {} \, dx =\int\limits^t_0 {-10\left ((t^3)/(3)-(7t^2)/(2)+10t \right ) +100t} \, dt\\ x=-10\left ((t^4)/(4)-(7t^3)/(6)+5t^2 \right )+50t^2+3`

a) At t =2 we have displacement

`x=-10\left ((2^4)/(4)-(7*2^3)/(6)+5\rimes2^2 \right )+50*2^2+3\\\\x=36.33\ \rm m`

b) Now Let v be the velocity at t=4 s then

`v=-10\left ((t^3)/(3)-(7t^2)/(2)+10t \right )\\\\v=-10\left ((4^3)/(3)-(7* 4^2)/(2)+10* 4\right )\\\\v=346.67\ \rm m/s`

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

`(dx)/(dt)=-10\left ((t^3)/(3)-(7t^2)/(2)+10t \right ) +100t=0\\t=0\ or\ 7`

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

`=(dv)/(dt)=10(-t+2)(t-5)+100=0\\\\t=7\ \rm s\\\\v_(max)=-10\left ((t^3)/(3)-(7t^2)/(2)+10t \right )\\\\v=-10\left ((7^3)/(3)-(7*7^2)/(2)+10* 7 \right )+100* 7=128.33\ \rm m/s`