Question

A car entering a highway stops on the entrance ramp. Then the car accelerates uniformly along a straight line, reaching 28 m/s in 5.6 s. (2 points)

A. What is the average acceleration of the car during the 5.6 s?

b. How far does the car travel in the 5.6 s?

Answer 1

According to the question,

Initial velocity of the car (
`\rm v_i`) = 0 m/s

Final velocity of the car (
`\rm v_f`) = 28 m/s

Time of uniform acceleration (t) = 5.6 s

a. By using
`1^(st)` kinematic equation and substituting value we get:

`\rm \implies v_f = v_i + at \\ \\ \rm \implies at = v_f - v_i \\ \\ \rm \implies a = (v_f - v_i )/(t) \\ \\ \rm \implies a = (28 - 0 )/(5.6) \\ \\ \rm \implies a = (28)/(5.6) \\ \\ \rm \implies a = 5 \: m {s}^( - 2)`

`\therefore` Average acceleration of the car during the 5.6 s is 5 m/s²

b. By using
`2^(st)` kinematic equation and substituting value we get:

`\rm \implies s = v_it + (1)/(2)at^2 \\ \\ \rm \implies s =0 * 5.6 + (1)/(2) * 5 * {(5.6)}^(2) \\ \\ \rm \implies s = (1)/(2) * 5 * 31.36 \\ \\ \rm \implies s =5 * 15.68 \\ \\ \rm \implies s =78.4 \: m`

`\therefore` Distance travelled by car in 5.6 s is 78.4 m