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 <em>x</em> be the displacement of her vehicle during this motion. Then
0² - (30.0 m/s)² = 2 (-8.00 m/s²) <em>x</em>
==> <em>x</em> = (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 <em>x</em> is the distance traveled as the car comes to a stop, then
0² - (21.0 m/s)² = 2 (-3.50 m/s²) <em>x</em>
==> <em>x</em> = (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 <em>a</em> such that
(11.5 m/s)² - 0² = 2 <em>a</em> (15.0 m)
==> <em>a</em> = (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)/<em>t</em>
==> <em>t</em> = (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 <em>v</em>, at its highest point we have
0² - <em>v</em> ² = 2 (-<em>g</em>) (91.5 m)
==> <em>v</em> = √(2<em>g</em> (91.5 m)) ≈ 42.3 m/s
(where I use <em>g</em> = 9.80 m/s²)
6.a. The brick's velocity after 7.0 s is
-<em>g</em> (7.0 s) = -68.6 m/s ≈ -69 m/s
6.b. The brick is presumably dropped from rest, so it is displaced by <em>x</em> such that
(-68.6 m/s)² - 0² = 2 (-<em>g</em>) <em>x</em>
==> <em>x</em> = -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.)
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