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A particle moves in a straight line with an initial velocity of 35 m/s and a constant acceleration of 38 m/s2. If at t = 0, x = 0, what is the particle's position (in m) at t = 6 s?

Respuesta :

Answer:

d=894 m

Explanation:

Given that

initial velocity ,u= 35 m/s

Acceleration ,a= 38 m/s²

time ,t= 6 s

Given that at t= 0 s ,x= 0 m

We know that

[tex]d=ut+\dfrac{1}{2}at^2 [/tex]

d=Displacement

Now by putting the values

[tex]d=35\times 6+\dfrac{1}{2}\times 38\times 6^2 [/tex]

d=894 m

Therefore the particle position after 6 sec will be 894 m.

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