Question

in 1997, the trust ssc, a turbofan powered vehicle, achieved the land speed world record of 760.34 mph (1,223.65 km/h) and became the first car to break the sound barrier. if it took 56 seconds for the trust ssc to reach its top speed, what was the average acceleration during that time?
a=?
t=56
vi=0
vf=1223.65
$\frac{km}{hr} cdot \frac{1000m}{1km} cdot \frac{1hr}{60 cdot 60s} = \frac{m}{s}$
$cdot \frac{1}{56} = \frac{m}{s^2}$

Explanation:

Step1: Convert speed to m/s

First, convert the final speed \( V_f = 1223.65\space \text{km/h} \) to \( \text{m/s} \). Using the conversion factors: \( \frac{\text{km}}{\text{h}} \cdot \frac{1000\space \text{m}}{1\space \text{km}} \cdot \frac{1\space \text{h}}{3600\space \text{s}} \). So, \( V_f = 1223.65\times\frac{1000}{3600}\space \text{m/s} \approx 339.90\space \text{m/s} \). Initial speed \( V_i = 0\space \text{m/s} \), time \( t = 56\space \text{s} \).

Step2: Use acceleration formula

The formula for average acceleration is \( a=\frac{V_f - V_i}{t} \). Substitute \( V_f \approx 339.90\space \text{m/s} \), \( V_i = 0\space \text{m/s} \), and \( t = 56\space \text{s} \) into the formula: \( a=\frac{339.90 - 0}{56}\approx 6.07\space \text{m/s}^2 \).

Answer:

The average acceleration is approximately \( 6.07\space \text{m/s}^2 \) (or more precisely, after exact calculation of \( 1223.65\times\frac{1000}{3600}\div56 \), we get \( \frac{1223650}{3600\times56}=\frac{1223650}{201600}\approx 6.07\space \text{m/s}^2 \))