Answer: 12.22 ft/sec²
Step-by-step explanation:
An increase from 26 to 51 is an increase of 51 - 26 = 25 mi/hr
We need to do this in 3 seconds --> 25 mi/hr ÷ 3 sec
Note the following conversion: 1 mile = 5280 ft
[tex]\dfrac{25\ miles}{hr}\times \dfrac{1}{3\ sec}\times \dfrac{5280\ ft}{1\ mile}\times \dfrac{1\ hr}{60\ min}\times \dfrac{1\ min}{60\ sec} \\\\\\=\dfrac{5280(25)\ ft}{3(60)(60)\ sec^2}\\\\\\=\large\boxed{12.22\ ft\slash sec^2}[/tex]
The constant acceleration that is required to increase the speed of a car from 26 mi/h to 51 mi/h in 3 seconds is 12.22 ft/s².
Acceleration can be defined as the rate of change of the velocity of an object with respect to time.
[tex]\rm Acceleration=\dfrac{Final\ velocity- Initial\ Velocity}{Time}[/tex]
As the velocity that is given to us is 51 miles/hour and 26 miles/hour, therefore, we first need to convert the units of the velocity in order to get the acceleration in ft/s².
[tex]\rm Final\ velocity= 51\ mi/hr = \dfrac{51\times 5280}{3600} = 74.8\ m\s^2[/tex]
[tex]\rm Initial\ velocity= 26\ mi/hr = \dfrac{26\times 5280}{3600} = 38.134\ m\s^2[/tex]
Now, acceleration is written as the ratio of the difference between the velocity and the time needed to increase or decrease the velocity of the object.
[tex]\rm Acceleration=\dfrac{Final\ velocity- Initial\ Velocity}{Time}[/tex]
Substituting the values we will get,
[tex]\rm Acceleration = \dfrac{74.8-38.134}{3} = 12.22\ \ ft/s^2[/tex]
Hence, the constant acceleration that is required to increase the speed of a car from 26 mi/h to 51 mi/h in 3 seconds is 12.22 ft/s².
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