Question

The speed of a particle moving on a straight path
is changed from 5 m/s to 10 m/s in covering a
distance of 5m. Its acceleration is​

Answer 1

7.5m/s^2

Explanation:

Given-

u=5m/s

v=10m/s

s=5m

a=?

from 3rd equation v^2- u^2= 2as

(10)^2 - (5)^2=2xaxs

100-25=10a

75=10a

a=75/10=7.5m/s^2

Answer 2

$\LARGE{\underline{\underline{\red{\sf{Answer :}}}}}$

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Initial Velocity (u) = 5 m/s

Final Velocity (v) = 10 m/s

Distance (s) = 5 m

We have equation of Motion :

$\Large{\underline{\boxed{\sf{v^2 \: - \: u^2 \: = \: 2as}}}}$

Put Values,

⇒(10)² - (5)² = 2(a)(5)

⇒100 - 25 = 10a

⇒75 = 10a

⇒a = 75/10

⇒a = 7.5

$\large{\boxed{\sf{Acceleration\: (a) \: = \: 7.5 \: ms^{-2}}}}$

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$\rule{200}{2}$

Additional Information

Proof for v² - u² = 2as

As we know,

$\LARGE \displaystyle \longrightarrow {\boxed{\sf{\overrightarrow{a} \: = \: \frac{dv}{dt}}}}$

Multiply Divide by dx,

$\Large \displaystyle \leadsto {\sf{\overrightarrow{a} \: = \: \frac{dv}{dt} \: \times \: \frac{dx}{dt}}}$

As dx/dt = v

$\Large \displaystyle \leadsto {\sf{\overrightarrow{a} \: = \: \frac{dv \: \times \: v}{dx}}}$

$\Large \displaystyle \leadsto {\sf{adx \: = \: dv^2}}$

Integrate both sides,

$\Large \displaystyle \leadsto {\sf{a \int dx \: = \: \int \frac{v^2}{2}}}$

$\Large \displaystyle \leadsto {\sf{a[x]^x_{x_{0}} \: = \: [\frac{v^2}{2}]^v_{u}}}$

$\Large \displaystyle \leadsto {\sf{a(x \: - \: x_{0}) \: = \: \frac{v^2 \: - \: u^2}{2}}}$

As (x - x0 = s)

$\Large \displaystyle \leadsto {\sf{a(s) \: = \: \frac{v^2 \: - \: u^2}{2}}}$

$\Large{\underline{\boxed{\red{\sf{v^2 \: - \: u^2 \: = \: 2as}}}}}$