Answer:
A. 25.08 s
B. 3082.53 m
C. 3×10⁵ m/s²
Explanation:
A. Determination of the time.
This can be obtained as illustrated below:
Acceleration (a) = –9.8 m/s²
Initial velocity (u) = 245.8 m/s
Final velocity (v) = 0 m/s
Time (t) =.?
v = u + at
0 = 245.8 + (–9.8 × t)
0 = 245.8 – 9.8t
Collect like terms
0 – 245.8 = – 9.8t
– 245.8 = – 9.8t
Divide both side by –9.8
t = –245.8 / –9.8
t = 25.08 s
Therefore, it will take 25.08 s for the car to come to a complete stop.
B. Determination of the distance travelled by the car.
Acceleration (a) = –9.8 m/s²
Initial velocity (u) = 245.8 m/s
Final velocity (v) = 0 m/s
Distance (s) =?
v² = u² + 2as
0² = 245.8² + (2 × –9.8 × s)
0 = 60417.64 – 19.6s
Collect like terms
0 – 60417.64 = – 19.6s
– 60417.64 = – 19.6s
Divide both side by –19.6
s = –60417.64 / –19.6
s = 3082.53 m
Thus, the car travelled a distance of 3082.53 m before stopping completely.
C. Determination of the acceleration of the object.
Initial velocity (u) = 0 m/s
Final velocity (v) = 600 m/s
Distance (s) = 0.6 m
Acceleration (a) =?
v² = u² + 2as
600² = 0² + (2 × a × 0.6)
360000 = 0 + 1.2a
360000 = 1.2a
Divide both side by 1.2
a = 360000 / 1.2
a = 300000 = 3×10⁵ m/s²
is the buoyant force
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is the acceleration due to gravity