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kifflom [539]
3 years ago
11

How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at

60 mph?
Mathematics
1 answer:
Aleks04 [339]3 years ago
8 0
The vehicle traveling 60 mph is going 4 times as fast. the brakes must do an amount of work that's equal to the original kinetic energy. the brakes exert the same amount of force no matter what the case is, so the stopping distance will be four times as long for the car going 60 mph compared to the car going 30 mph.
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Consider the differential equationy′′+3y′−10y=0.(a) Find the general solution to this differential equation.(b) Now solve the in
RSB [31]

Answer:

Y= 2e^(5t)

Step-by-step explanation:

Taking Laplace of the given differential equation:

s^2+3s-10=0

s^2+5s-2s-10=0

s(s+5)-2(s+5) =0

(s-2) (s+5) =0

s=2, s=-5

Hence, the general solution will be:

Y=Ae^(-2t)+ Be^(5t)………………………………(D)

Put t = 0 in equation (D)

Y (0) =A+B

2 =A+B……………………………………… (i)

Now take derivative of (D) with respect to "t", we get:

Y=-2Ae^(-2t)+5Be^(5t)   ....................... (E)

Put t = 0 in equation (E) we get:

Y’ (0) = -2A+5B

10  = -2A+5B ……………………………………(ii)

2(i) + (ii) =>

2A+2B=4 .....................(iii)

-2A+5B=10 .................(iv)

Solving (iii) and (iv)

7B=14

B=2

Now put B=2 in (i)

A=2-2

A=0

By putting the values of A and B in equation (D)

Y= 2e^(5t)

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