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Nady [450]
3 years ago
8

A car is traveling with an initial velocity of 15 m/s when the driver decides to go faster by accelerating at 5 m/s2 for 10 seco

nds. How far did the car travel during this time?
70 m


300 m


175 m


400 m
Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0

Answer: Time needed: 2.5 s

Distance covered: 31.3 m

Explanation:

I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by

 

on one side of the equation and solve by plugging your values

To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation

Explanation:

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Answer : The initial rate of the reaction is, 1.739\times 10^{-3}s^{-1}

Explanation :

First we have to calculate the rate constant of the reaction.

Expression for rate law for first order kinetics is given by :

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant  = ?

t = time taken for the process  = 44 s

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[A] = amount or concentration left time 44 s = 0.0554 M

Now put all the given values in above equation, we get:

k=\frac{2.303}{44}\log\frac{0.1108}{0.0554}

k=0.0157

Now we have to calculate the initial rate of the reaction.

Initial rate = K [A]

At t = 0, [A]=[A_o]

Initial rate = 0.0157 × 0.1108 = 1.739\times 10^{-3}s^{-1}

Therefore, the initial rate of the reaction is, 1.739\times 10^{-3}s^{-1}

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Learn more:

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