The useful equations here are derived equations for rectilinear motion at constant acceleration:
a = (v - v₀)/t
2ax = v² - v₀²
x = v₀t + 0.5at²
1. Since it starts from rest, v₀ = 0. Then, v = 10.5 m/s and t=1.35 s
a = (10.5 - 0)/1.35 = 7.78 m/s²
a = 7.78 m/s²*(1 km/1000 m)*(3,600 s/ 1 h)² = 100,828.8 km/h²
2. v₀ = 0; v = 60 km/h; t=5.4 s
a = [60 km/h*(1000 m/1km)*(1 h/3600 s) - 0]/5.4 s = 3.09 m/s²
3. Using the acceleration and velocities in #2, we can determine the distance by the formula: 2ax = v² - v₀²
2(3.09 m/s²)x = [60 km/h*(1000 m/1km)*(1 h/3600 s)]² - (0 m/s)²
Solving for x,
x = 44.95 m
4. Using acceleration in #2 and v₀ = 0, the time would be
x = v₀t + 0.5at²
0.25 km * 1,000 m/1 km = (0)(t) + (0.5)(3.09 m/s²)(t²)
Solving for t,
t = 12.72 seconds
5. Acceleration of automobile:
a = [(55 km/h - 25 km/h)*(1,000 m/1 km)*(1 h/3,600 s)]/30 s
a = 0.278 m/s²
Acceleration of bicycle:
a = [(30 km/h - 0 km/h)*(1,000 m/1 km)*(1 h/3,600 s)]/(0.5 min * 60 s/1 min)
a = 0.278 m/s²
6. x = 85 km or 85,000 m; v₀ = 0 m/s; v = 2,800 m/s
2ax = v² - v₀²
2a(85,000 m) = (2,800 m/s)² - 0²
a = 46.12 m/s²
7. Using the data given and acceleration in #6,
x = v₀t + 0.5at²
85,000 m = 0*t + 0.5(46.12 m/s²)(t²)
Solving for t,
t = 60.71 seconds