A car, initially at rest, accelerates at a constant rate of 2 m/s^2 due east (ch

Question

A car, initially at rest, accelerates at a constant rate of 2 m/s^2 due east (choose east as the positive direction: round your answer to the nearest meter per second: include units) What is its velocity after 1 second? _____________ What is its velocity after 2 seconds? _____________ What is its velocity after 3 seconds? _____________ What is its velocity after 4 seconds? _____________ What is its velocity after 5 seconds? _____________ What is its velocity after 9 seconds? _____________ If instead a car accelerates at a constant rate of 5 m/s^2 due east What is its velocity after 1 second? _____________ What is its velocity after 2 seconds? _____________ What is its velocity after 3 seconds? _____________ If instead a car accelerates at a constant rate of 5 m/s^2 due west what is its velocity after 1 second? _____________ A car, initially at rest, accelerates at a constant rate of 2 m/s^2 due east (choose east as the positive direction: round your answer to the nearest meter per second: include units). What is its average velocity during the 1^st second? _____________ What is its average velocity during the 2^nd second? _____________ What is its average velocity during the 3^rd second? _____________ What is its average velocity during the 4^th second? _____________ What is its average velocity during the 5^th second? _____________ What is its average velocity during the 9^th second? _____________ If instead the car accelerates at 5 m/s^2 due east (round your answer to the tenth of a meter per second). What is its average velocity during the 1^st second? _____________ What is its average velocity during the 2nd second? _____________

Explanation / Answer

Question 11]

Use the equation of motion, v = u + at

where v = final velocity at time t , u = initial velocity and a = acceleration

here, u = 0 m/s and a = 2m/s2

therefore, at t = 1s, v = 2 m/s

at t = 2s, v = 0 + (2 x 2) = 4 m/s

at t = 3s, v = 6 m/s

at t = 4s, v = 8 m/s

at t = 5s, v = 10 m/s and at t = 9s. v = 0 + (2x9) = 18 m/s.

If instead, a = 5 m/s2

then , at t = 1s, v = 0 + 5 = 5 m/s

at t = 2s, v = 2 x 5 = 10 m/s

and at t = 3s, v = 3 x 5 = 15 m/s

If car now accelerates in the opposite direction with acceleration a = 5m/s2,

then, at t = 1s : v = 0 - (1 x 5) = - 5m/s.

12] a = 2 m/s2

At t = 1s, v = 2 m/s

so vavr = (2 + 0 )/2 = 1 m/s

at t = 2s, v = 0 + (2 x 2) = 4 m/s

so, vavr = (0 + 4)/2 = 2 m/s

at t = 3s, v = 0 + (2 x 3) = 6 m/s

therefore, vavr = (6 + 0)/2 = 3m/s.

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