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CGP EDU Academic Team
Published on: September 12, 2026
Two cars A and B at rest at same point initially. If A starts with uniform velocity of 40 m/sec and B starts in the same direction with constant acceleration of \(4 \mathrm{m} / \mathrm{s}^2\) , then B will catch A after how much time
Text Solution
Verified by ExpertsThe correct answer is:
B
Let A and B will meet after time t sec . it means the distance travelled by both will be equal.
\(S_A \quad \text{wt} \quad 40 \Omega\) and \(S_y - \frac{1}{2} a t^2 = \frac{1}{2} \times 4 \times t^2\)
\(s_a - s_e = 40t - \frac{1}{2}4t^2 = t - 20\) sec
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