- #1

- 274

- 0

I have no clue how to solve this.

I used the usual method without integration and got s=1700m. The textbook ans says 1690.

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- MHB
- Thread starter Shah 72
- Start date

- #1

- 274

- 0

I have no clue how to solve this.

I used the usual method without integration and got s=1700m. The textbook ans says 1690.

- #2

- 1,103

- 1

$d=1690$m to three significant figures

- #3

- 274

- 0

Thank you!

$d=1690$m to three significant figures

- #4

- 274

- 0

$d=1690$m to three significant figures

Can I pls ask you about the last part, how you got 4.5 ( v156+v165) do I need to integrate?

- #5

- 274

- 0

I don't understand the last part.

$d=1690$m to three significant figures

I get s1 for the interval between 0 and 6= 25.92m

Constant velocity, v=s/t, I get s2 for interval between 6 and 156= 1620m.

- #6

- 1,103

- 1

recall the kinematics equation for uniform acceleration …

$\Delta x = v_{avg} \cdot \Delta t = \dfrac{1}{2}(v_0 + v_f) \cdot \Delta t = 4.5(10.8 + 0) = 48.6$ m

you could also use …

$\Delta x = v_0 \Delta t + \dfrac{1}{2}at^2 = 10.8 \cdot 9 - 0.6 \cdot 9^2 = 48.6$ m

- #7

- 274

- 0

Thanks a lot!uniformdeceleration, $\Delta t = 9$ sec.

recall the kinematics equation for uniform acceleration …

$\Delta x = v_{avg} \cdot \Delta t = \dfrac{1}{2}(v_0 + v_f) \cdot \Delta t = 4.5(10.8 + 0) = 48.6$ m

you could also use …

$\Delta x = v_0 \Delta t + \dfrac{1}{2}at^2 = 10.8 \cdot 9 - 0.6 \cdot 9^2 = 48.6$ m

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