- #1

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- Thread starter SarahV
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- #1

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- #2

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aceeleration=change in velocity/change in time so:

17-8/3=a

d=at

so d=a * 3 seconds

17-8/3=a

d=at

so d=a * 3 seconds

- #3

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Actually, d=at is incorrect.SarahV said:

The correct formula is [tex] v^2 - u^2 = 2as [/tex]. u is the initial velocity, in this case 8 m/s. We know [tex] a = 3 m/s^2 [/tex], therefore [tex] 17^2 - 8^2 = 2(3)s [/tex] and so [tex] s = (17^2 - 8^2)/6 [/tex] which turns out to be 37.5 m.

- #4

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8+3a=17

a=3m/s^2

8*3+0.5*3*3^2=24+13.5=37.5m

a=3m/s^2

8*3+0.5*3*3^2=24+13.5=37.5m

- #5

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Let me summarize the solution.

a=acceleration u=initial velocity v=final velocity

s=displacement t=time taken

a=(v-u)/t

a=(17.0-8.0)/3.0

a=3 m/(s^2)

s=((u+v)*t)/2

s=((8.0+17.0)*3.0)/2

s=37.5 m

Therefore, the acceleration is 3 m/(s^2) to his original direction and

the displacement he travelled is 37.5 m to his original direction.

a=acceleration u=initial velocity v=final velocity

s=displacement t=time taken

a=(v-u)/t

a=(17.0-8.0)/3.0

a=3 m/(s^2)

s=((u+v)*t)/2

s=((8.0+17.0)*3.0)/2

s=37.5 m

Therefore, the acceleration is 3 m/(s^2) to his original direction and

the displacement he travelled is 37.5 m to his original direction.

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