Solve a constant-acceleration motion problem
Use s = ut + ½at² and v² = u² + 2as to find distance and final speed.
For steady acceleration, three formulas cover most problems: v = u + at, s = ut + ½at², and v² = u² + 2as. Here u is the starting speed, v the final speed, a the acceleration, t the time and s the distance.
Example: a cart starts at 5 m/s and speeds up at 2 m/s² for 4 s. Distance: s = 5 × 4 + ½ × 2 × 4² = 20 + 16 = 36 m. Then the final speed from v² = 25 + 2 × 2 × 36 = 169, so v = 13 m/s.
Classic Graphing 84 · Distance
Start on the main screen: Press 2nd then mode (QUIT) to get back to the home screen. Pressing clear on a menu also backs out of it.
Type u × t + ½ × a × t² with the numbers. Use .5 for one half.
5×4+.5×2×4x²Press enter. The distance is 36 m.
enter
You should see5*4+.5*2*4² 36
Classic Graphing 84 · Final speed
Start on the main screen: Press 2nd then mode (QUIT) to get back to the home screen. Pressing clear on a menu also backs out of it.
Use the square root key (2nd, x²). Type √(5² + 2 × 2 × 36) and close the bracket.
2nd√x²5x²+2×2×36)Press enter. The final speed is 13 m/s.
enter
You should see√(5²+2*2*36) 13
Natural Scientific 991 · Distance
Start on the main screen: Press MENU, then 1 (Calculate) to return to the main calculation screen. AC clears the line you are on.
Type 5 × 4 + .5 × 2 × 4² and press = .
5×4+.5×2×4x²=
You should see5×4+.5×2×4² 36
Natural Scientific 991 · Final speed
Start on the main screen: Press MENU, then 1 (Calculate) to return to the main calculation screen. AC clears the line you are on.
Press the square-root key and type 5² + 2 × 2 × 36. The root sign stays open until you close it. Press = .
√5x²+2×2×36=
You should see13
Tips
- List what you know (u, a, t) and what you want (s, v), and choose the formula that has exactly those letters.
- Moving up against gravity, a is negative: a = -9.8 m/s².