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joja [24]
4 years ago
14

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

by it in this time
Physics
1 answer:
Soloha48 [4]4 years ago
6 0

Answer:

\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}

\boxed{\sf Distance \ covered \ (s) = 100 \ m}

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

\sf (i) \ From \ 1^{st} \ equation \ of \ motion:

\sf \implies v = u + at

\sf \implies 20 = 0 + a(10)

\sf \implies 10a = 20

\sf \implies \frac{10a}{10}  =  \frac{20}{10}

\sf \implies a = 2 \: m/ {s}^{2}

\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:

\sf \implies s = ut +  \frac{1}{2} a {t}^{2}

\sf \implies s = (0)(10) +  \frac{1}{2}  \times 2 \times  {(10)}^{2}

\sf \implies s =  \frac{1}{ \cancel{2}}  \times  \cancel{2} \times  {(10)}^{2}

\sf \implies s =  {10}^{2}

\sf \implies s = 100 \: m

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Now we can define the electric potential energy and the electric potential difference:

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The electric potential energy is also defined as the work done on a charge q moved through a potential difference of \Delta V. Consequently, the potential difference \Delta V represents the work per unit charge done, i.e. the work done when moving a unitary charge through a potential difference \Delta V.

Learn more about potential difference and current:

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