First slide

The 3 equation for uniform acc

Question

A particle starts from origin with a velocity of 10 m/s and continues to move along positive x-axis with uniform retardation of 5 m/s2. Then length of path traversed by the particle when it cross the origin again is

Easy

A
30 m
B
20 m
C
60 m
D
40 m

Solution

When the particle stops momentarily , its displacement x = $\frac{10^{2}}{2 \times 5} m = 10 m$
Therefore length of path traversed = 2 $\times 10 m = 20 m$

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