Question

When a 1984 Alfa Romeo Spider sports car accelerates at the maximum possible rate, its motion during the first 20 is extremely well modeled by the simple equation v2x=2Pmt where P=3.6104W is the car's power output, m = 1200kg is its mass, and vx is in m/s. That is, the square of the car's velocity increases linearly with time.

a) What is the car's speed at t=11 s ?

b) What is the car's speed at t=18 s ?

c)Find a symbolic expression, in terms of P, m, and t, for the car's acceleration at time t.

d)Evaluate the acceleration at t=1.0 s .

e) Evaluate the acceleration at t=10 s .

Answer 1

a-)
`v_(x) = 25.69 m/s`

b-)
`v_(x) = 32.86 m/s`

c-)
`a = \frac{\sqrt{ (2P)/(m)}}{ 2√(t)}}\\`

d-)
`a= 3.87 m/s^(2)`

e-)
`a= 1.22 m/s^(2)`

Step-by-step explanation:

First, lay out the given data:

P = 3.6 x 10^4 W

m=1200 kg

`v_(x) ^(2)=(2P)/(m)t\\v_(x) ^(2)=(2*3.6*10^(4))/(1200) *t`

a)What is the car's speed at t=11 s ?

`v_(x)^(2)=(2*3.6*10^(4))/(1200) *t\\v_(x)=√(60*11) \\v_(x) = 25.69 m/s`

b) What is the car's speed at t=18 s ?

`v_(x)^(2)=(2*3.6*10^(4))/(1200) *t\\v_(x)=√(60*18) \\v_(x) = 32.86 m/s`

c) Find a symbolic expression, in terms of P, m, and t, for the car's acceleration at time t.

An expression for the acceleration can be defined by finding the derivative of the velocity equation:

`v_(x) =\sqrt{ (2P)/(m)t}\\(d(v_(x)))/(dt) = (d)/(dt)(\sqrt{ (2P)/(m)t})\\(d(v_(x)))/(dt) = (d)/(dt)(\sqrt{ (2P)/(m)}*t^{(1)/(2)} )\\(d(v_(x)))/(dt) = (\sqrt{ (2P)/(m)}* (1)/(2)t^{(-1)/(2)} )\\\\(d(v_(x)))/(dt) = \frac{\sqrt{ (2P)/(m)}}{ 2√(t)}}\\a = \frac{\sqrt{ (2P)/(m)}}{ 2√(t)}}\\`

d) Evaluate the acceleration at t=1.0 s

`a = \frac{\sqrt{ (2P)/(m)}}{ 2√(t)}}\\\\a = \frac{\sqrt{ (2*3.6*10^(4))/(1200)}}{ 2√(1)}}\\\\a= 3.87 m/s^(2)`

e) Evaluate the acceleration at t=10 s

`a = \frac{\sqrt{ (2P)/(m)}}{ 2√(t)}}\\\\a = \frac{\sqrt{ (2*3.6*10^(4))/(1200)}}{ 2√(10)}}\\\\a= 1.22 m/s^(2)`