Let's work through Acceleration — Complete Guide with Examples together. Physics problems become simple when you break them into clear, logical steps. Here's exactly how to approach this one.
See the step-by-step solution above for the complete answer.
Step 1: a = (vf - vi)/t
Step 2: a = (6 - 0)/2 = 6/2 = 3 m/s²
Step 3: a = (vf - vi)/t
Step 4: a = (6 - 0)/2 = 6/2 = 3 m/s²
Step 5: a = (vf - vi)/t
Step 6: a = (11 - 2)/3 = 9/3 = 3 m/s²
Step 7: a = (vf - vi)/t
Step 8: a = (16 - 4)/4 = 12/4 = 3 m/s²
Step 9: a = (vf - vi)/t
Step 10: a = (21 - 6)/5 = 15/5 = 3 m/s²
Step 11: a = (vf - vi)/t
Step 12: a = (26 - 8)/6 = 18/6 = 3 m/s²
Step 13: a = (vf - vi)/t
Step 14: a = (31 - 10)/7 = 21/7 = 3 m/s²
Step 15: a = (vf - vi)/t
Step 16: a = (36 - 12)/2 = 24/2 = 12 m/s²
Step 17: a = (vf - vi)/t
Step 18: a = (41 - 14)/3 = 27/3 = 9 m/s²
Step 19: a = (vf - vi)/t
Step 20: a = (22 - 16)/4 = 6/4 = 1.5 m/s²
Step 21: a = (vf - vi)/t
Step 22: a = (27 - 18)/5 = 9/5 = 1.8 m/s²
Step 23: a = (vf - vi)/t
Step 24: a = (12 - 0)/6 = 12/6 = 2 m/s²
Step 25: a = (vf - vi)/t
Step 26: a = (17 - 2)/7 = 15/7 = 2.142857142857143 m/s²
Step 27: a = (vf - vi)/t
Step 28: a = (22 - 4)/2 = 18/2 = 9 m/s²
Step 29: a = (vf - vi)/t
Step 30: a = (27 - 6)/3 = 21/3 = 7 m/s²
Step 31: a = (vf - vi)/t
Step 32: a = (32 - 8)/4 = 24/4 = 6 m/s²
Step 33: a = (vf - vi)/t
Step 34: a = (37 - 10)/5 = 27/5 = 5.4 m/s²
Step 35: a = (vf - vi)/t
Step 36: a = (18 - 12)/6 = 6/6 = 1 m/s²
Step 37: a = (vf - vi)/t
Step 38: a = (23 - 14)/7 = 9/7 = 1.2857142857142858 m/s²
Step 39: a = (vf - vi)/t
Step 40: a = (28 - 16)/2 = 12/2 = 6 m/s²
Step 41: a = (vf - vi)/t
Step 42: a = (33 - 18)/3 = 15/3 = 5 m/s²
This problem applies acceleration. Understanding the underlying principle lets you solve similar problems with different values confidently.
❌ Mistake 1: Using the wrong units or forgetting to convert.
❌ Mistake 2: Applying the wrong formula for the given situation.
❌ Mistake 3: Rounding too early in the calculation.
See the step-by-step solution above for the complete answer.
This problem applies acceleration. Understanding the underlying principle lets you solve similar problems with different values confidently.