Physics, 10.07.2019 02:30 braphi5609

Acar accelerates from 14 m/s to 21m/s in 6.0 s. what was its acceleration? how far did it travel in this time? assume constant acceleration.

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Answer from: wilkvictor07

The acceleration of the car is 1.17 m/s^2

The car traveled 104.70 m in 6 seconds with constant acceleration.

Explanation:

Initial speed of the car = u = 14 m/s

Final speed of the car = v = 21 m/s

Acceleration of the car in in 6 seconds = a

t = time = 6 seconds

Using first equation of motion:

v=u+at

$21 m/s=14 m/s+a\times6s$

$7 m/s=a\times 6s$

$a=\frac{7 m/s}{6 s}=1.1667 m/s^2\approx 1.17m/s^2$

The acceleration of the car is 1.17 m/s^2

Distance traveled by the car in 6 seconds: = s

Using third equation of motion:

v^2-u^2=2as

$(21 m/s)^2-(14 m/s)^2=2\times 1.2 m/s^2\times s$

$s=\frac{(21 m/s)^2-(14 m/s)^2}{2\times 1.17 m/s^2}=104.70 m$

The car traveled 104.70 m in 6 seconds with constant acceleration.

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Acar accelerates from 14 m/s to 21m/s in 6.0 s. what was its acceleration? how far did it travel in...

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