Question

a car accelerates from 6m/s to 18m/s in 12 seconds. Assuming acceleration to be uniform calculate a) the acceleration & b) the distance covered in 12seconds.

Answer 1

Here,
Initial speed, u = 6 m/s
Final speed, v = 18 m/s
Time, t = 12 s

We have to find,
Acceleration, a = ?
Distance, s = ?

Now, by first equation of motion, we know that,

$v = u + at$
$= > \: a = \frac{v - u}{t}$
Putting values of u, v and t from the question, we get
$= > a = \frac{18 - 6}{12}$
$= > a = \frac{12}{12}$
$= > a = 1$
Therefore, acceleration of the car =
1 m/s²

Now, by Second equation of motion, we know that,

$s = ut + \frac{1}{2} a {t}^{2}$
Putting the values of u, t and a, we get

$= > s =(6 \times 12) + (\frac{1}{2} \times 1 \times {12}^{2} )$
$= > s = 72 + ( \frac{1}{2} \times 1 \times 144)$
$= > s = 72 + ( \frac{1}{2} \times 144)$
$= > s =72 + ( \frac{144}{2} )$
$= > s = 72 + 72$
$= > s = 144$

Therefore, distance travelled by the car in 12 sec = 144 m.

Hope it'll help.. :-D

Answer 2

Answer:

144 m

hope this helps you