Answer:
5.266 secs
Step-by-step explanation:
Lets assume ; p(t) = t^-3 + 2^2 + ( 3/2 ) is the particle position along x-axis
time interval [ 0, 4 ]
Average velocity = Displacement / time
= p( b ) - p( a ) / b - a -------- ( 1 )
where a = 0 , b = 4 ( time intervals )
Back to equation 1
Average velocity = [ ( 4^-3 + 4 + (3/2) ) - ( 0 + 4 + (3/2) ) ] / 4
= 3.9 * 10^-3 ----- ( 2 )
Instantaneous velocity = d/dx p(t)
= - 3/t^4 ------ ( 3 )
To determine the time that the instanteous velocity = average velocity
equate equations (2) and (3)
3.9*10^-3 = - 3 / t^4
t^4 = - 3 / ( 3.9 * 10^-3 ) = - 769.231
hence t = ![\sqrt[4]{- 769.231 }](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B-%20769.231%20%7D)
= 5.266 secs
we ignore the negative sign because time can not be in the negative
Answer: #4 x >l 11
Hope this helps
Step-by-step explanation:
hi
it'b : If you earn money from a coupon, it's mean that you lentmoney to somenone.in In stocks, there is no coupon. The annual meeting of stock holders will decide how much is given for each share by considering annual profit made by the compagny.
I disagree. If each side of the equation is divided by 5, the result is x2 = 4. By the square root property of equality, x = -2 or x = 2. So x could be -2 instead of 2.
Answer:
speed=d/t
let him cover x distance while walking
running d=4.5-x
s1=x/0.5 km/hr....(1)
s2=4.5-x//5/12.....(2)
atq.....s2=2+s1
s1=2x
s2=54-12x/5
by eqn 1 and 2
s1+s2=2x+54-12x/5
putting value of s2
s1+2+s1=2x+54-12x/5....taking LCM =5
10s1+10= 10x+54-12x
as s1=2x
10s1+10= -2x+54
10s1+10=-s1+54
11s1=44
speed when walking=4km/hr
