Question

An object moves in a straight path. Its position x as a function of time t is presented by the equation x(t) = at – bt2+c, where a = 1.4 m/s, b = 0.06 m/s2 and c =50 m. a. Calculate the average velocity of the object for the time interval t = 0 to t = 5 s b. Calculate the average velocity of the object for the time interval t = 0 to t = 10 s c. Calculate the average velocity of the object for the time interval t = 10 to t = 15 s

Answer 1

Answer:

The answer is below

Explanation:

Given that:

x(t) = at – bt2+c

a) x(t) = at – bt2+c

Substituting a = 1.4 m/s, b = 0.06 m/s2 and c =50 m gives:

x(t) = 1.4t - 0.06t² + 50

At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

$v=\frac{x(5)-x(0)}{5-0}\\ \\v=\frac{55.5-50}{5-0}=1.1\\ \\v=1.1\ m/s$

b) x(t) = 1.4t - 0.06t² + 50

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

$v=\frac{x(10)-x(0)}{10-0}\\ \\v=\frac{58-50}{10-0}=0.8\\ \\v=0.8\ m/s$

c) x(t) = 1.4t - 0.06t² + 50

At t = 15, x(5) = 1.4(15) - 0.06(15)² + 50 = 57.5 m

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

The average velocity (v) is given as:

$v=\frac{x(15)-x(10)}{15-10}\\ \\v=\frac{57.5-58}{15-10}=0.1\\ \\v=0.1\ m/s$