Hw2-2 - show all your work, including the equation, express all answers in base units, (m, m/s, m/s2), unless stated otherwise i
n problem. a sprinter accelerates from rest to 10.5 m/s in 1.35 sec. what is her acceleration . . . in m/s2 km/hr2 a certain drag racer can accelerate from 0 to 60.0 km/hr in 5.4 sec. what is its acceleration in m/s2? how far, in meters, will it travel during this acceleration? how much time would it take to go 0.25 km if it could accelerate at this rate the entire time? an automobile can increase its speed uniformly from 25.0 to 55.0 km/h in 30.0 seconds. a bicycle rider uniformly speeds up to 30.0 km/h from rest in 0.5 min. calculate their accelerations. a rocket rises from rest with constant acceleration to an altitude of 85.0 km, at this point it has a speed of 2.8 km/s what is the acceleration of the rocket? how long does the ascent take?
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 = <span>10.5 m/s and t=1.35 s a = (10.5 - 0)/1.35 = 7.78 m/s</span>² 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 <span> </span>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 <span> 6. x = 85 km or 85,000 m; v</span>₀ = 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
The force of the car increases because it is proportional to the acceleration of the object (as stated in the second law). If the object accelerates then the force accelerates, if the acceleration decreases then the force decreases. Same for mass. The force of the car would have changed from 315 newtons of force to 1,260 newtons of force.
The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre. For example, the density of water is 1 gram per cubic centimetre, and Earth's density is 5.51 grams per cubic centimetre.
