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A jeeprey acceleration of a speed of
40m//s over a distance of
300m. Determine the aceleration (assume uniformly) af the jeeprey.

$1$ . A jeeprey acceleration of a speed of $40 \mathrm{~m} / \mathrm{s}$ over a distance of $300 \mathrm{~m}$ . Determine the aceleration (assume uniformly) af the jeeprey.

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q.

1

. A jeeprey acceleration of a speed of

40 \mathrm{~m} / \mathrm{s}

over a distance of

300 \mathrm{~m}

. Determine the aceleration (assume uniformly) af the jeeprey.

Use acceleration formula: We need to use the formula for acceleration, which is $a = \frac{v^2 - u^2}{2 \cdot s}$ , where $v$ is the final velocity, $u$ is the initial velocity (which we assume to be $0$ since it's acceleration from rest), and $s$ is the distance.
Plug in known values: First, let's plug in the values we know: $v = 40\,\text{m/s}$ , $u = 0\,\text{m/s}$ (since it's starting from rest), and $s = 300\,\text{m}$ .
Calculate acceleration: Now, calculate the acceleration using the formula: $a = \frac{{(40^2 - 0^2)}}{{(2 \times 300)}}$ .
Simplify equation: Simplify the equation: $a = \frac{1600 - 0}{600}$ .
Perform math: Do the math: $a = \frac{1600}{600}$ .
Find acceleration: Finally, we get the acceleration: $a = 2.67 \, \text{m/s}^2$ .

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