Answer:
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.
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 [tex]=\frac{5.10}{1.95}=2.62m/s[/tex]
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 [tex]=\frac{19.48}{3.96}=4.92m/s[/tex]
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 [tex]=\frac{19.48-5.10}{2.01}=7.15m/s[/tex]
