Question

A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t)= α t2− β t3, where α = 1.44 m/s2 and β = 5.00×10−2 m/s3

a. Calculate the average velocity of the car for the time interval t=0 to t1 = 1.95 s .

b. Calculate the average velocity of the car for the time interval t=0 to t2 = 3.96 s .

c. Calculate the average velocity of the car for the time interval t1 = 1.95 s to t2 = 3.96 s .

Answer 1

a) Average velocity of the car for the time interval t=0 to t = 1.95 s is 2.62 m/s.

b) Average velocity of the car for the time interval t=0 to t = 3.96 s is 4.92 m/s.

c) Average velocity of the car for the time interval t=1.95 to t = 3.96 s is 7.15 m/s.

Step-by-step explanation:

We have x(t)= α t²− β t³

That is x(t)= 1.44 t²− 5 x 10⁻² t³

Average velocity is ratio of distance traveled to time.

a)Average velocity of the car for the time interval t=0 to t = 1.95 s

x (0) = 1.44 x 0²− 5 x 10⁻² x 0³ = 0 m

x (1.95) = 1.44 x 1.95²− 5 x 10⁻² x 1.95³ = 5.10 m

Time difference = 1.95 - 0 = 1.95 s

Average velocity
`=(5.10)/(1.95)=2.62m/s`

b)Average velocity of the car for the time interval t=0 to t = 3.96 s

x (0) = 1.44 x 0²− 5 x 10⁻² x 0³ = 0 m

x (3.96) = 1.44 x 3.96²− 5 x 10⁻² x 3.96³ = 19.48 m

Time difference = 3.96 - 0 = 3.96 s

Average velocity
`=(19.48)/(3.96)=4.92m/s`

c)Average velocity of the car for the time interval t=0 to t = 3.96 s

x (1.95) = 1.44 x 1.95²− 5 x 10⁻² x 1.95³ = 5.10 m

x (3.96) = 1.44 x 3.96²− 5 x 10⁻² x 3.96³ = 19.48 m

Time difference = 3.96 - 1.95 = 2.01 s

Average velocity
`=(19.48-5.10)/(2.01)=7.15m/s`