A car travels 100 m while decelerating to 8 m/s in 5 s. initial speed is 32

b) What was it's acceleration?

Vi = 32 [m/s]

Explanation:

In order to solve this problem we must use the following the two following kinematics equations.

The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.

where:

Vf = final velocity = 8[m/s]

Vi = initial velocity [m/s]

a = acceleration = [m/s^2]

t = time = 5 [s]

Now replacing:

8 = Vi - 5*a

Vi = (8 + 5*a)

As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.

where:

d = distance = 100[m]

(8^2) = (8 + 5*a)^2 - (2*a*100)

64 = (64 + 80*a + 25*a^2) - 200*a

0 = 80*a - 200*a + 25*a^2

0 = - 120*a + 25*a^2

0 = 25*a(a - 4.8)

therefore:

a = 0 or a = 4.8 [m/s^2]

We choose the value of 4.8 as the acceleration value, since the zero value would not apply.

Returning to the first equation:

8 = Vi - (4.8*5)

Vi = 32 [m/s]

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adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation: