Physics, asked by JnvdsKushal6126, 8 months ago

1.A cyclist driving at 5 m/s, picksup velocity of 10 m/s over a distance of 15 m.Calculate acceleration and time in which the cyclist picksup above velocity.

Answers

Answered by Anonymous
37

GiveN :

  • Initial velocity (u) = 5 m/s
  • Final velocity (v) = 10 m/s
  • Distance travelled (s) = 15 m

To FinD :

  • Acceleration of cyclist
  • Time interval in which it attains velocity of 10 m/s

SolutioN :

Use 3rd equation of motion :

\implies v² - = 2as

ㅤㅤㅤㅤㅤㅤㅤ

\implies 10² - 5² = 2 * a * 15

ㅤㅤㅤㅤㅤㅤㅤ

\implies 100 - 25 = 30a

ㅤㅤㅤㅤㅤㅤㅤ

\implies 75 = 30a

ㅤㅤㅤㅤㅤㅤㅤ

\implies a = 75/30

ㅤㅤㅤㅤㅤㅤㅤ

\implies a = 2.5

ㅤㅤㅤㅤㅤㅤㅤ

\underline{\sf{\therefore \: Acceleration\ is\ 2.5\ ms^{-2}}}

________________________

Now, use 1st equation of motion :

\implies v = u + at

ㅤㅤㅤㅤㅤㅤㅤ

\implies 10 = 5 + 2.5 * t

ㅤㅤㅤㅤㅤㅤㅤ

\implies 10 - 5 = 2.5t

ㅤㅤㅤㅤㅤㅤㅤ

\implies 2.5t = 5

ㅤㅤㅤㅤㅤㅤㅤ

\implies t = 5/2.5

ㅤㅤㅤㅤㅤㅤㅤ

\implies t = 2 s

ㅤㅤㅤㅤㅤㅤㅤ

\underline{\sf{\therefore \: Time\ taken\ by\ cyclist\ is\ 2s}}


Abhishek474241: Perfect
Answered by MaIeficent
59

Explanation:

{\red{\underline{\underline{\bold{Given:-}}}}}

  • Initial velocity(u) = 5m/s

  • Final velocity(v) = 10m/s

  • Distance covered(s) = 15m

{\blue{\underline{\underline{\bold{To\:Find:-}}}}}

  • The acceleration and time in which the cyclist picks up the velocity.

{\green{\underline{\underline{\bold{Solution:-}}}}}

According to the third equation of motion

\boxed{ \sf \pink{ \rightarrow {v}^{2} =  {u}^{2}  + 2as }}

Substituting the values:-

\sf \rightarrow {10}^{2} =  {5}^{2}  + 2(a)(15)

\sf \rightarrow {100}=  {25}  + 30a

\sf \rightarrow 30a =  100-25

 \sf \rightarrow 30a =  75

\sf \rightarrow a =   \dfrac{75}{30}

\sf \rightarrow a =  2.5

According to the first equation of motion:-

\boxed{ \sf \orange{ \rightarrow {v}=  {u}  + at }}

Substituting the values

{ \sf { \rightarrow {10}=  {5}  + 2.5(t )}}

{ \sf { \rightarrow {2.5t}=  10 - 5}}

{ \sf { \rightarrow {2.5t}=   5}}

{ \sf { \rightarrow {t}=    \dfrac{5}{2.5} }}

{ \sf { \rightarrow {t}=    2}}

Hence;

  \boxed{ \sf \purple{Acceleration \:  = 2.5m/ {s}^{2} }}

  \boxed{ \sf \purple{ \rightarrow Time \:  = 2sec}}

Similar questions