Question

A particle initially at rest moves with a uniform acceleration of 10 m\s-2 and attains a velocity of 90m\s after some time. the time taken to travel a distance of 405m is​

Answer 1

$\mathfrak{\huge{\underline{Answer:}}}$

GIVEN:

Initially at rest [U]= 0

Uniform acceleration [a]= 10 m/s^-2

Velocity after 't' time= 90 m/s

Displacement [S]= 405 m

To find: Time taken to travel the displacement;

According to Equation of Motion;

S=ut+½at²

405=0*t+½at²

405*2=10t²

810/10=t²

t=√81

t=9

The particles takes 9 sec to travel the distance of 405m.

HOPE THIS WILL HELP YOU....

Answer 2

$\huge{\sf{===Answer===}}$

Initial Velocity (u) = 0

Acceleration (a) = 10 m/s²

Final Velocity (v) = 90 m/s

Distance (s) = 405 m

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We have formula :-

$\huge{\boxed{\boxed{\sf{S \: = \: ut \: + \: \frac{1}{2} \: at^2}}}}$

405 = 0(t) + 1 * 10(t²)/2

405 = 10t²/2

405 = 5t²

t² = 405/5

t² = 81

t = √81

t = 9

$\huge{\boxed{\boxed{\sf{Time \: = \: 9 \: s}}}}$