Question

A sprinter accelerates from rest to 9.00 in 1.38 sec.What is his acceleration in a. m/s2b. m/h2

Answer 1

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 $9.00\,\frac{m}{s}$ in 1.38 seconds. What is his acceleration in a. $\frac{m}{s^{2}}$ and b. $\frac{km}{h^{2}}$?

a) We assume that sprinter accelerates uniformly, so that acceleration ($a$), measured in meters per square second, can be obtained from this kinematic expression as function of initial and final velocities ($v_{o}$, $v$), measured in meters per second, and time ($t$), measured in seconds, as well.

$v = v_{o} + a\cdot t$

$a = \frac{v-v_{o}}{t}$

If we know that $v_{o} = 0\,\frac{m}{s}$, $v = 9\,\frac{m}{s}$ and $t = 1.38\,s$, the acceleration experimented by the sprinter is:

$a = \frac{9\,\frac{m}{s}-0\,\frac{m}{s} }{1.38\,s}$

$a = 6.521\,\frac{m}{s^{2}}$

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:

$a = \left(6.521\,\frac{m}{s^{2}} \right)\cdot \left(\frac{1}{1000}\,\frac{km}{m} \right)\cdot \left(3600^{2}\,\frac{s^{2}}{h^{2}} \right)$

$a = 84512.16\,\frac{km}{h^{2}}$

The acceleration of 84512.16 kilometers per square hour.