1.Consider the position of a ball thrown down with an initial speed of 25 m/s. W

Question

1.Consider the position of a ball thrown down
with an initial speed of 25 m/s.
What will be its position after 3.3 s? Let
the initial position be 0. The acceleration of
gravity is 9.8 m/s^2.Answer in units of m.

2.Consider the position of a ball thrown up
with an initial speed of 25 m/s.What will be its position after 3.3 s?
Answer in units of m.

3.A car, moving along a straight stretch of highway, begins to accelerate at 0.021 m/s^2. It takes the car 78.6 s to cover 1 km. How fast was the car going when it ?rst began to accelerate? Answer in units of m/s

Explanation / Answer

1.Consider the position of a ball thrown down with an initial speed of 25 m/s. What will be its position after 3.3 s? Let the initial position be 0. The acceleration of gravity is 9.8 m/s^2.Answer in units of m. X = XO + VO*T + 0.5AT^2 X = 0 +25m/s*3.3 s + 0.5*9.8 m/s^2*(3.3 s)^2 X = 0 + 82.5 m + 53.4 m X = 135.9 m 2.Consider the position of a ball thrown up with an initial speed of 25 m/s.What will be its position after 3.3 s? Answer in units of m. X = XO + VO*T + 0.5AT^2 X = 0 + 25 m/s*3.3 s + 0.5(-9.8 m/s^2)*(3.3 s)^2 X = 0 + 82.5 m – 53.4 m X = 29.1 m 3.A car, moving along a straight stretch of highway, begins to accelerate at 0.021 m/s^2. It takes the car 78.6 s to cover 1 km. How fast was the car going when it ?rst began to accelerate? Answer in units of m/s X = XO + VO*T + 0.5AT^2 1 km = 0 + VO*78.6 s + 0.5*0.021 m/s^2*(78.6 s)^2 1000 m = VO*78.6 s + 64.9 m 935.1 m = VO*78.6 s Initial velocity = 935.1 m/78.6 s Initial velocity = 11.9 m/s

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