Question

A car 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.

Answer 1

Explanation:

It is given that,

Initial velocity of the car, u = 14 m/s

Final velocity of the car, v = 21 m/s

Time taken, t = 6 s

Acceleration of the car is equal to the change in speed divided by time i.e. it can be written as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{21\ m/s-14\ m/s}{6\ s}$  

a = 1.166 m/s²

or

a = 1.17 m/s²

For finding how fat it travel in this time is calculated using third equation of motion :

$v^2-u^2=2as$

s = distance

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{21^2-14^2}{2\times 1.17}$

s = 104.7 m

Hence, the acceleration of the car is 1.166 m/s² and distance covered by it is 104.7 m

Answer 2

Answer:

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.