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A cheetah accelerating uniformly from rest reaches a speed of
29ms^(-1) in
2.0s and then maintains this speed for 15 seconds.
(a) Calculate
(i) its acceleration
(ii) the distance it travels while accelerating

A cheetah accelerating uniformly from rest reaches a speed of $29\,\text{ms}^{-1}$ in $2.0\,\text{s}$ and then maintains this speed for $15\,\text{seconds}$ . (a) Calculate (i) its acceleration (ii) the distance it travels while accelerating

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. A cheetah accelerating uniformly from rest reaches a speed of

29\,\text{ms}^{-1}

in

2.0\,\text{s}

and then maintains this speed for

15\,\text{seconds}

. (a) Calculate (i) its acceleration (ii) the distance it travels while accelerating

Find acceleration formula: To find the acceleration, use the formula $a = \frac{v - u}{t}$ , where $v$ is the final velocity, $u$ is the initial velocity, and $t$ is the time taken.
Calculate acceleration: Plug in the values: $a = \frac{29 \, \text{m/s} - 0 \, \text{m/s}}{2.0 \, \text{s}} = \frac{29}{2} \, \text{m/s}^2$ .
Calculate distance while accelerating: Calculate the acceleration: $a = 14.5 \, \text{m/s}^2$ .
Plug in values for distance calculation: Now, calculate the distance traveled while accelerating using the formula $s = ut + \frac{1}{2}at^2$ , where $s$ is the distance, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.
Calculate distance: Plug in the values: $s = 0 \, \text{m/s} \times 2.0 \, \text{s} + \frac{1}{2} \times 14.5 \, \text{m/s}^2 \times (2.0 \, \text{s})^2$ .
Calculate distance: Plug in the values: $s = 0 \, \text{m/s} \times 2.0 \, \text{s} + \frac{1}{2} \times 14.5 \, \text{m/s}^2 \times (2.0 \, \text{s})^2$ . Calculate the distance: $s = 0 + \frac{1}{2} \times 14.5 \times 4 = 29 \, \text{m}$ .

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