EXAM · MOTION IN A STRAIGHT LINE
OVERTAKING WITH ACCELERATION
Car B has a 50 m head start and moves at a constant 15 m/s. Car A starts from rest at the same moment and accelerates at 3 m/s². At what time does Car A catch Car B, and how fast is Car A moving at that instant?
§ 01
Step-by-step solution
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Step 1
Setting x_A = x_B gives a quadratic in t. Compute the discriminant Δ = v_B² + 2·a_A·d_0.
Hint
The quadratic ½·a_A·t² − v_B·t − d_0 = 0 has discriminant b² − 4ac; identify a = ½·a_A, b = −v_B, c = −d_0.
Step 2
Solve for the positive root: t_equal = (v_B + √Δ) / a_A.
Step 3
Substitute t_equal into x_A = ½·a_A·t² to find the position of the overtake.
Step 4
Find Car A's speed at the moment of overtaking: v_A = a_A · t_equal.
Solution walkthrough
This problem requires setting two position functions equal and solving the resulting quadratic. Car A starts from rest: x_A = ½·a_A·t² = 1.5t². Car B has a head start: x_B = d_0 + v_B·t = 50 + 15t. Setting x_A = x_B: 1.5t² = 50 + 15t, rearranged to 1.5t² − 15t − 50 = 0. Apply the quadratic formula with a = 1.5, b = −15, c = −50: discriminant = b² − 4ac = 225 + 300 = 525. The positive root is t = (15 + √525) / 3 ≈ 12.64 s. At that time, x_overtake = 1.5 × (12.64)² ≈ 239.6 m, and Car A's speed is a_A × t_equal = 3 × 12.64 ≈ 37.91 m/s. Notice that Car A is travelling more than twice as fast as Car B at the moment of overtaking — it needed that speed advantage to overcome the 50 m deficit. The cross-check x_B = 50 + 15 × 12.64 ≈ 239.6 m confirms both trains are at the same position, as required.
§ 02
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