Answer:
a) The acceleration of the sprinter is 6.521 meters per square second, b) The acceleration of 84512.16 kilometers per square hour.
Explanation:
Note: Statement is incomplete, complete description of the problem is: A sprinter accelerates from rest to [tex]9.00\,\frac{m}{s}[/tex] in 1.38 seconds. What is his acceleration in a. [tex]\frac{m}{s^{2}}[/tex] and b. [tex]\frac{km}{h^{2}}[/tex]?
a) We assume that sprinter accelerates uniformly, so that acceleration ([tex]a[/tex]), measured in meters per square second, can be obtained from this kinematic expression as function of initial and final velocities ([tex]v_{o}[/tex], [tex]v[/tex]), measured in meters per second, and time ([tex]t[/tex]), measured in seconds, as well.
[tex]v = v_{o} + a\cdot t[/tex]
[tex]a = \frac{v-v_{o}}{t}[/tex]
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 9\,\frac{m}{s}[/tex] and [tex]t = 1.38\,s[/tex], the acceleration experimented by the sprinter is:
[tex]a = \frac{9\,\frac{m}{s}-0\,\frac{m}{s} }{1.38\,s}[/tex]
[tex]a = 6.521\,\frac{m}{s^{2}}[/tex]
The acceleration of the sprinter is 6.521 meters per square second.
b) If we know that a hour is equivalent to 3600 seconds and a kilometer is equivalent to 1000 meters, the result of the previous item is converted by using two consecutive unit conversions:
[tex]a = \left(6.521\,\frac{m}{s^{2}} \right)\cdot \left(\frac{1}{1000}\,\frac{km}{m} \right)\cdot \left(3600^{2}\,\frac{s^{2}}{h^{2}} \right)[/tex]
[tex]a = 84512.16\,\frac{km}{h^{2}}[/tex]
The acceleration of 84512.16 kilometers per square hour.
