Answer:
A) (1 s, 2.3 s)
B) (-4 m/s², 3.8 m/s²)
Step-by-step explanation:
The car's position which is the distance is given by the equation;
s(t) = t³ - 5t² + 7t
A) Velocity is the first derivative of the distance. Thus;
v(t) = ds/dt = 3t² - 10t + 7
At v = 0, we have;
3t² - 10t + 7 = 0
Using quadratic formula, we have;
t = 1 and t = 2.3
Thus, time at velocity of 0 is t = (1 s, 2.3 s)
B) acceleration is the derivative of the velocity. Thus;
a(t) = dV/dt = 6t - 10
At velocity of 0, we got t = 1 and t = 2.3
Thus;
a(1) = 6(1) - 10 = -4 m/s²
a(2.3) = 6(2.3) - 10 = 3.8 m/s
Thus, a(t) at v = 0 gives; (-4 m/s², 3.8 m/s²)
Answer:
damm its tuff to not know anything
Step-by-step explanation:
Answer:
b. 9/100
Step-by-step explanation:
It helps if you have memorized the cubes of small integers. Of course, the ones needed here are ...
3³ = 27
10³ = 1000
The desired value can be simplified to ...
![\left(-\dfrac{1000}{27}\right)^{-2/3}=\left(-\sqrt[3]{\dfrac{3^3}{10^3}}\right)^2=\dfrac{(-3)^2}{10^2}=\dfrac{9}{100}](https://tex.z-dn.net/?f=%5Cleft%28-%5Cdfrac%7B1000%7D%7B27%7D%5Cright%29%5E%7B-2%2F3%7D%3D%5Cleft%28-%5Csqrt%5B3%5D%7B%5Cdfrac%7B3%5E3%7D%7B10%5E3%7D%7D%5Cright%29%5E2%3D%5Cdfrac%7B%28-3%29%5E2%7D%7B10%5E2%7D%3D%5Cdfrac%7B9%7D%7B100%7D)
You would use the = sign because
8+7-2^2=3^2+2
they both equal 11
