Question

a cheetah can accelerate from rest to a speed of 21.0 m/s in 6.75 s. what is its acceleration (in m/s²)?

Explanation:

Step1: Recall acceleration formula

Acceleration \( a \) is defined as the change in velocity \( \Delta v \) over time \( \Delta t \), so \( a=\frac{\Delta v}{\Delta t} \). The cheetah starts from rest, so initial velocity \( v_0 = 0\space m/s \), final velocity \( v = 21.0\space m/s \), time \( t = 6.75\space s \). Then \( \Delta v=v - v_0=21.0 - 0 = 21.0\space m/s \).

Step2: Calculate acceleration

Substitute \( \Delta v = 21.0\space m/s \) and \( \Delta t=6.75\space s \) into the formula: \( a=\frac{21.0}{6.75}\space m/s^2 \). Calculating this, \( \frac{21.0}{6.75}=\frac{2100}{675}=\frac{28}{9}\approx 3.148\space m/s^2 \) (or more precisely, \( \frac{21.0}{6.75} = 3.111\cdots\approx3.11\space m/s^2 \) if we keep more decimal precision in calculation, but let's do exact division: \( 21\div6.75 = 21\div\frac{27}{4}=21\times\frac{4}{27}=\frac{28}{9}\approx 3.11\space m/s^2 \) (wait, \( 21\div6.75 = 3.111...\), actually \( 6.75\times3 = 20.25 \), \( 21 - 20.25 = 0.75 \), \( 0.75\div6.75=\frac{1}{9}\approx0.111 \), so total is \( 3.111...\approx3.11\space m/s^2 \) or \( 3.148 \) was a miscalculation. Wait, \( 21\div6.75 \): multiply numerator and denominator by 100 to get \( 2100\div675 \), divide numerator and denominator by 75: \( 28\div9\approx3.11\space m/s^2 \) (since \( 9\times3.11 = 27.99 \), close to 28).

Answer:

\( \approx 3.11 \) (or more accurately, \( \frac{28}{9}\approx3.11 \) or using calculator \( 21\div6.75 = 3.111\cdots\approx3.11\space m/s^2 \))