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A particle initially moving East with a speed of 20.0 m/s, experiences an acceleration of 3.95 m/s, North for a time of 8.00 s. What was the speed of the particle after this acceleration, in units of m/s? Give the answer as a positive number.

Respuesta :

Answer:

Speed of particle, v = 51.6 m/s

Explanation:

It is given that,

Initial speed of the particle which is moving towards east, u = 20 m/s

Acceleration of the particle, a = 3.95 m/s²

Time taken, t = 8 s

We have to find the speed of the particle after this acceleration. Let the speed is v. It can be calculated using first equation of motion as :

v = u + at

[tex]v=20\ m/s+3.95\ m/s^2\times 8\ s[/tex]

v = 51.6 m/s

So, the speed of the particle after this acceleration is 51.6 m/s

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