I have no idea how to do this question. Would be appreciated if someone help =)
The velocity of a car is given by v(t)=60(1-[0.7]^t), where v is measured in m/s and t is measured in s. Determine the following:
a)What is the acceleration function?
b) What is the accerlation at t=2s?
c)What is the initial velocity?
d)What is the time at which the accleration is 3m/s^2?
I assume you have calculus.
a= dv/dt= d(60-60*.7t)/dt=-60*.7^t * ln0.7
Yes, I do have calculus.
So the acceleration function is -60(0.7t)(In0.7)
b) What is the accerlation at t=2s?
-60(0.7(2))(In0.7)= 29.996.
The acceleration would be 29.99 m/s^2?
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How do I do part c and d?
Sure! Let's break down each part of the question:
a) To find the acceleration function, we need to take the derivative of the velocity function with respect to time, which represents the rate of change of velocity.
The given velocity function is v(t) = 60(1 - [0.7]^t). To find the derivative, we can use the power rule and chain rule of differentiation.
Differentiating 60 with respect to t gives us 0 since it is a constant. Then, differentiating (1 - [0.7]^t) with respect to t, we get -ln(0.7) * [0.7]^t.
Therefore, the acceleration function, a(t), is -ln(0.7) * [0.7]^t.
b) To find the acceleration at t = 2s, we can substitute t = 2 into the acceleration function we found in part a).
a(2) = -ln(0.7) * [0.7]^2 = -ln(0.7) * (0.49) ≈ -0.357 * 0.49 ≈ -0.175 m/s^2.
So, the acceleration at t = 2s is approximately -0.175 m/s^2.
c) The initial velocity represents the velocity when t = 0. To find this, we substitute t = 0 into the velocity function.
v(0) = 60(1 - [0.7]^0) = 60(1 - 1) = 60 * 0 = 0 m/s.
Therefore, the initial velocity is 0 m/s.
d) To find the time at which the acceleration is 3 m/s^2, we need to solve the equation a(t) = 3.
-ln(0.7) * [0.7]^t = 3.
To solve this equation, we can first divide both sides by -ln(0.7):
[0.7]^t = -3/ln(0.7).
We can then take the logarithm base 0.7 on both sides to solve for t:
t * log(0.7) = log(-3/ln(0.7)).
Finally, divide both sides by log(0.7):
t = log(-3/ln(0.7)) / log(0.7).
Note that the number inside the logarithm is negative, which means there is no real solution for t. Therefore, there is no time at which the acceleration is 3 m/s^2 in the given scenario.
I hope this explanation helps! Let me know if you have any more questions.