Question

An auto’s velocity decreases uniformly from 20m/s to 6m/s while covering 70m. The acceleration
is

Answer 1

Given:

An auto’s velocity decreases uniformly from 20m/s to 6m/s while covering 70m.  

To find:

Find the acceleration

Solution:

From given, we have,

An auto’s velocity decreases uniformly from 20m/s to 6m/s while covering 70m.  

⇒ s = 70 m, u = 20 m/s, v = 6 m/s

we use the formula:

s = ut + 1/2 at²

upon differentiating the above equation, we get,

⇒ v = u + at

substituting the given values in the above equation, we get,

⇒ 6 = 20 + at

∴ at = -14

Now again consider,

s = ut + 1/2 at²

⇒ s = ut + 1/2 (at) × t

⇒ 70 = 20t + 1/2 (-14) × t

⇒ 70 = 20t - 7t

⇒ 70 = 13t

∴ t = 70/13 s.

we have,

at = -14

a (70/13) = -14

∴ a = -2.6 m/s²

The negative sign indicates the opposite direction.

Therefore, the acceleration of an auto is -2.6 m/s²

Answer 2

Given that,

Initial speed of auto, u = 20 m/s

Final speed of auto, v = 6 m/s

Distance, d = 70 m

To find,

Acceleration of auto.

Solution,

Using second equation of motion,

$d=ut+\dfrac{1}{2}at^2\\\\70=20t+\dfrac{1}{2}at^2\ .....(1)$

Using first equation of motion,

$v=u+at\\\\6=20+at\\\\at=-14$......(2)

Put at=-14 in equation (1)

So,

$70=20t+\dfrac{1}{2}(at)t\\\\70=20t+\dfrac{1}{2}(-14)t\\\\70=20t-7t\\\\t=\dfrac{70}{13}\ s$

Put the value of t in equation (2). So,

$at=-14\\\\a=\dfrac{-14}{t}\\\\a=\dfrac{-14\times 13}{70}\\\\a=-2.6\ m/s^2$

So, the acceleration of the auto is $-2.6\ m/s^2$.