Question

Please help!!!!!!!!!

Answer 1

1. This question asks about velocity, so A and B are not correct. The car's velocity after 15 s with acceleration 2.00 m/s² would be

(2.00 m/s²) (15 s) = 30 m/s

[D]

2. Because Ima is slowing down to a stop, the acceleration is negative. Let x be the displacement of her vehicle during this motion. Then

0² - (30.0 m/s)² = 2 (-8.00 m/s²) x

==>   x = (30.0 m/s)²/(2 (8.00 m/s²)) = 56.25 m ≈ 65.3 m

[A]

3. Since acceleration is constant, the average velocity is exactly the average of the initial and final velocities:

(21.0 m/s + 0 m/s)/2 = 10.5 m/s

The average (and thus instantaneous) acceleration during this time is equal to the change in velocity divided by the change in time:

(0 m/s - 21.0 m/s)/(6.00 s) = -3.50 m/s²

If x is the distance traveled as the car comes to a stop, then

0² - (21.0 m/s)² = 2 (-3.50 m/s²) x

==>   x = (21.0 m/s)² / (2 (3.50 m/s²)) = 63.0 m

[A]

4.a. Assuming the sprinter's acceleration is constant, the average acceleration would be a such that

(11.5 m/s)² - 0² = 2 a (15.0 m)

==>   a = (11.5 m/s)² / (2 (15.0 m)) ≈ 4.41 m/s²

4.b. By definition of average acceleration,

4.41 m/s² = (11.5 m/s - 0 m/s)/t

==>   t = (11.5 m/s)/(4.41 m/s²) ≈ 2.61 s

5. At maximum height, any thrown object has zero velocity, so if it was thrown with an initial speed v, at its highest point we have

0² - v ² = 2 (-g) (91.5 m)

==>   v = √(2g (91.5 m)) ≈ 42.3 m/s

(where I use g = 9.80 m/s²)

6.a. The brick's velocity after 7.0 s is

-g (7.0 s) = -68.6 m/s ≈ -69 m/s

6.b. The brick is presumably dropped from rest, so it is displaced by x such that

(-68.6 m/s)² - 0² = 2 (-g) x

==>   x = -240.1 m ≈ -240 m

which is to say it falls a distance of 240 m. (The displacement is negative because we take its initial position to be the origin, and I took the downward direction to be negative.)