A zebra starts from rest and accelerates at $1.9$ $1.9$ $m$ $m$ $/ s^{2}$ $/s^2$ . How far has the zebra gone after $5$ $5$ seconds?

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A zebra starts from rest and accelerates at $1.9$ $1.9$ $m$ $m$ $/ s^{2}$ $/s^2$ . How far has the zebra gone after $5$ $5$ seconds?

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Answer 1 · 2016-12-10T15:31:40

Using formula dist = $\frac{1}{2}$ $1/2$ * a * $t^{2}$ $t^2$ , the zebra has gone 23.75 meters

Explanation:

At the end of the first second the zebra is going at 1.9 m/s.
At the end of the second second the zebra is going at 3.8 m/s
...
At the end of the fifth second the zebra is going at ( $5 \cdot 1.9 =$ $5 * 1.9 =$ ) 9.5 m/s
The zebra's Average Speed was ( $\frac{9.5}{2} =$ $9.5 / 2 =$ ) 4.75 m/s
Thus in 5 seconds the zebra has travelled ( $5 \cdot 4.75 =$ $5 * 4.75 =$ ) 23.75 meters

The basic kinematics formulas to solve such problems is dist = $\frac{1}{2}$ $1/2$ * a * $t^{2}$ $t^2$

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A zebra starts from rest and accelerates at $1.9$ $1.9$ $m$ $m$ $/ s^{2}$ $/s^2$ . How far has the zebra gone after $5$ $5$ seconds?

1 Answer

Explanation:

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A zebra starts from rest and accelerates at $1.9$ $1.9$ $m$ $m$ $/ s^{2}$ $/s^2$ . How far has the zebra gone after $5$ $5$ seconds?