What constant acceleration is required to increase the speed of a car from 26 mi/h to 56 mi/h in 5 s?

What constant acceleration is required to increase the speed of a car from 26 mi/h to 56 mi/h in 5 s? (Round your answer to two decimal places.)

_____________ft/s^2

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Use the formula v=u+at where v is the final velocity, u is the initial velocity, a is acceleration and t is time

t = 5 seconds

v = 56 mph = 56*5280/3600 f/s

u = 26mph = 26*5280/3600 f/s

the formula can be reduced to a = (v-u)/t

a = (56*5280/3600-26*5280/3600)/5 = 8.80 f/s^2

**Acceleration of car is 8.80 f/s^2**

The value of the acceleration of the car over a duration of 5 seconds in terms of ft/s^2 to increase the speed from 26 miles/hr to 56 miles/hr has to be determined.

First convert the given speeds to ft/s.

26 miles/hr = 38.13 ft/s and 56 miles/hr = 82.13 ft/s.

The value of the acceleration is (82.12 - 38.13)/5 = 44/5 = 8.8 ft/s^2

**An acceleration of 8.8 ft/s^2 is required to increase the speed from 26 mi/hr to 56 mi/hr in 5 s.**

Start by stating your necessary unit conversions:

`1 hr = 3600 s`

`1mi=5280 ft`

The required formula to calculate acceleration is:

`a = (v2 - v1) / t`

where, `a = `acceleration, `v2 = ` final speed, `v1 = ` initial speed, and `t = time`

` ``v1 = 26` mi/h `* 1/3600`h/s `* 5280` ft/mi

`= 38.13 ft/s`

`v2 = 56 ` mi/h `*1/3600` h/s `*5280` ft/mi

`= 82.13 ft/s`

and finally, we know that `t = 5s`

We substitute our values and solve:

`a = (82.13-38.13)/5`

`= 8.80` `ft/s^2`

**`:.` The constant acceleration will be `8.80 ft/s^2.` **

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