Question

An auto's velocity increases uniformly from 6.0 m/s to 20 m/s while covering 70 m in a straight line. Find the acceleration and the time taken.

Answer 1

answer:

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

Explanation:

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Answer 2

Given : Auto's velocity increases uniformly from 6.0 m/s to 20 m/s.

            Distance covered is 70 m.                

To Find : Acceleration and Time taken.

Solution :

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

As it is increased uniformly then,

From Equation of motion,

$\[{v^2} - {u^2} = 2as\]$

where,

u= initial velocity = 6m/s

v= final velocity = 20m/s

s= Distance covered = 70 m.

From given equation,

$\[{20^2} - {6^2} = 2 \times a \times 70\]$

$\[a = \frac{{400 - 36}}{{70 \times 2}}\]$

a = 2.6 m/$\[{s^2}\]$

Acceleration of auto is 2.6 m/$\[{s^2}\]$.

Now,

for Time taken

From Equation of Motion,

$\[a = \frac{{v - u}}{t}\]$

putting value in equation

$\[2.6 = \frac{{20 - 6}}{t}\]$

t=5.4 sec

Hence, Acceleration of the auto is 2.6 m/$\[{s^2}\]$ and time taken is 5.4 sec.