A car's position is given by s(t) = {3 – 5t? + 7t hundreds of meters with t in minutes. a. Find the time when the car has velocity zero.
b. Find the acceleration of the car when the velocity is zero.
Separate your answers with commas; round all values to 1 decimal(s).
Time(s) at which the velocity is zero:
Acceleration(s) when the velocity is zero:

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)

50 min

step-by-step explanation:

they will be decreasing both their times in half

60 / 2 = 30

40/2 = 20

30 + 20 = 50

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints