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 = 1.60 m/s2 and β = 5.10×10−2 m/s3

Answer 1

Given info : 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 = 1.60 m/s² and β = 5.10 × 10¯² m/s³

To find : Calculate the average velocity of the car for each time interval: (a) t = 0 to t = 2.00 s; (b) t = 0 to t = 4.00 s; (c) t = 2.00 s to t = 4.00 s.

solution : average velocity is ratio of the total displacement to total time taken by particle.

so, $\quad v(t)=\frac{x(t_2)-x(t_1)}{t_2-t_1}$

(a) average velocity = $\frac{x(2)-x(0}{2-1}$

= [1.6 × (2)² - 5 × 10¯² × 2³ - 0]/2

= 1.6 × 2 - 20 × 10¯²

= 3.2 - 0.2

= 3 m/s

(b) average velocity = $\frac{x(4)-x(0)}{4-0}$

= [1.6 × 4² - 5 × 10¯² × 4³]/4

= 1.6 × 4 - 80 × 10¯²

= 6.4 - 0.8

= 5.6 m/s

(c) average velocity = $\frac{x(4)-x(2)}{4-2}$

= [1.6 × (4)² - 5 × 10¯² × 4³ - 1.6 × 2² + 5 × 10¯² × 2³ ]/2

= 12.8 - 160 × 10¯² - 3.2 + 20 × 10¯²

= 12.8 - 1.6 - 3.2 + 0.2

= 13 - 4.8

= 8.2 m/s