Question

If an arrow is shot upward on Mars with a speed of 56 m/s, its height in meters t seconds later is given by y = 56t − 1.86t^2. Find the average speed over the given time intervals. [1, 2]

Answer 1

Answer:

50.42m/s

Step-by-step explanation:

Average speed is the change in distance of a body with respect to time. This is expressed as:

$v = \delta y/\delta t\\v = \dfrac{y(t_2)-y(t_1)}{t_2-t_1}$

Given the function of the height in metres expressed as y = 56t − 1.86t² and time interval [1,2]

t₁ = 1sec and t₂ = 2 secs

y(t₂) = 56t₂ − 1.86t₂²

y(2) = 56(2) − 1.86(2)²

y(2) = 112-7.44

y(2) = 104.56

y(t₁ ) = 56t₁  − 1.86t₁ ²

y(1) = 56(1) − 1.86(1)²

y(1) = 56-1.86

y(1) = 54.14

Substituting the derived value into the formula for finding the average speed:

$v = \dfrac{y(t_2)-y(t_1)}{t_2-t_1}\\\\v = \dfrac{104.56-54.14}{2-1}\\\\v = \dfrac{50.42}{1}\\\\v = 50.42m/s$

Hence, the average speed over the given time intervals. [1, 2] is 50.42m/s