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.)
3 Answers | Add Yours
Start by stating your necessary unit conversions:
`1 hr = 3600 s`
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.`
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.
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
We’ve answered 330,341 questions. We can answer yours, too.Ask a question