Question:
Two cars, P and Q move toward each other, getting \(8\ \text{m}\) closer every second. If they move in the same direction, they get \(0.6\ \text{m}\) closer every second. If car P is faster than car Q then, the ratio of speeds of car P to car Q is
Two cars, P and Q move toward each other, getting \(8\ \text{m}\) closer every second. If they move in the same direction, they get \(0.6\ \text{m}\) closer every second. If car P is faster than car Q then, the ratio of speeds of car P to car Q is
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Moving towards each other the relative speed is the sum of the speeds; moving in the same direction it is the difference. Solve the two equations.
Updated On: Nov 12, 2026
- \(43:37\)
- \(37:43\)
- \(74:43\)
- \(74:37\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
The rate at which the gap changes is the relative speed. Towards each other it is \(v_P+v_Q\), and in the same direction it is \(v_P-v_Q\) since P is faster.
Step 2: Write the equations
\(v_P+v_Q=8\ \text{m/s}\) and \(v_P-v_Q=0.6\ \text{m/s}\).
Step 3: Calculation
Adding gives \(2v_P=8.6\), so \(v_P=4.3\ \text{m/s}\). Subtracting gives \(2v_Q=7.4\), so \(v_Q=3.7\ \text{m/s}\).
Step 4: Take the ratio
\(v_P:v_Q=4.3:3.7=43:37\).
Final Answer:
The ratio of the speed of P to the speed of Q is 43:37. \[ \boxed{\text{(A) }43:37} \]
The rate at which the gap changes is the relative speed. Towards each other it is \(v_P+v_Q\), and in the same direction it is \(v_P-v_Q\) since P is faster.
Step 2: Write the equations
\(v_P+v_Q=8\ \text{m/s}\) and \(v_P-v_Q=0.6\ \text{m/s}\).
Step 3: Calculation
Adding gives \(2v_P=8.6\), so \(v_P=4.3\ \text{m/s}\). Subtracting gives \(2v_Q=7.4\), so \(v_Q=3.7\ \text{m/s}\).
Step 4: Take the ratio
\(v_P:v_Q=4.3:3.7=43:37\).
Final Answer:
The ratio of the speed of P to the speed of Q is 43:37. \[ \boxed{\text{(A) }43:37} \]
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