Question

the veolicity of a particle at an instant is 10 m/s after 5sec their velocity in 20m/s the velocity at 3 sec before is​

Answer 1

$\textbf{ANSWER}$

$\red{\textbf{intial velocity 'u' }}$

= 10 m/s

$\red{\textbf{final velocity 'v'}}$= 20m/s

$\red{\textbf{time taken }}$= 5 sec

$\textbf{Acc. to the first equation of motion}$

$\blue{\textbf{ v= u +at }}$

$\texttt{it has taken 5 seconds to accelerate by 10m/s}$

$\texttt{so ,10/5}$

= $\textbf{2m/s^2 }$

$\texttt{so assuming that the acceleration remains constant}$

$\texttt{according to the question}$,

$\textbf{for the instant 3 sec before }$

$\texttt{v= 10 m/s}$

$\texttt{t= 3 sec}$

$\texttt{using, }$

$\textbf{V = u + at}$

$\texttt{u= v - at }$

$\texttt{ 10- 2 *3 }$

= $\textbf{4m/s}$

$\blue{\textbf{ hence, the velocity was 4m/s }}$

Answer 2

Initial velocity = 10 m/sec

Final velocity = 20 m/sec

Time taken = 5 sec

Now we know that the rate in the change in velocity is defined as acceleration.

So, acceleration = Final velocity - initial velocity / time

= 20-10 /5

=10/5

=2 m/sec^2

Now, from the first equation of motion

v=u+at

Given that v =10 m/sec

a =2 m/sec^2 (calculated above)

t=3 sec

u=?

v=u+at

10 = u +2 ×3

10 = u +6

10-6 = u

4 m/sec = u

So, the velocity 3 seconds before was 4m/sec