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Numerical · Q20

Q.Two cars move towards each other along the same straight road, one with a velocity of 20 m/s and the other with a velocity of 15 m/s. Find the velocity with which the two cars approach each other.

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Setting up. Take the direction of motion of the first car as positive. Then car A has velocity vA=+20v_A = +20 m/s, and car B, moving toward A along the same line (i.e. in the opposite direction), has velocity vB=−15v_B = -15 m/s.

Relative velocity of A with respect to B. Using the definition from Section 2.9,

vAB=vA−vB=20−(−15)=20+15=35 m/sv_{AB} = v_A - v_B = 20 - (-15) = 20 + 15 = 35\ \text{m/s}

Physical interpretation. The magnitude of this relative velocity, 35 m/s, is the rate at which the distance separating the two cars is closing — i.e. the rate of approach. Notice this is the SUM of the two individual speeds (20 m/s and 15 m/s), not their difference, precisely because the cars move in opposite directions: subtracting a negative velocity is the same as adding its magnitude. This is a general feature — whenever two bodies on the same straight lin …

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