NOTE: You must ENTER THE PROPER UNITS in the answer box and your answer should b

Question

NOTE: You must ENTER THE PROPER UNITS in the answer box and your answer should be given to THREE SIGNIFICANT FIGURES.

A particle located initially at -6.00 m accelerates from rest to a velocity of 2.50 m/s in a time interval of 1.20 s.

a) Calculate the acceleration of the particle.

HINTS: What is the initial velocity if the particle starts out "at rest"? Don't make this part harder than it needs to be. Use only the information you need to find the acceleration.

b) Calculate the distance the particle moves during the 1.20 s time interval.

HINTS: Notice that this problem asks for the distance travelled, which is really the displacement, x. If you use xi in your calculation, then xf will be the final position, and not the distance traveled.

Explanation / Answer

here,

location of particle is -6 m

initial velocity , u = 0

final velocity , v = 2.5 m/s

time interval , t = 1.2 s

(a)

let the accelration of the particle be a

using first equation of motion

v = u + a * t

2.5 = 0 + a * 1.2

a = 2.083 m/s^2

the accelration of the particle is 2.083 m/s^2

(b)

let the distance travelled be s

using third equation of motion

v^2 - u^2 = 2 * a * s

2.5^2 - 0 = 2 * 2.083 * s

s = 1.500 m

the distance travelled in time intreval of 1.20 s is 1.500 m

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