The variation of the velocity of a particle moving along a straight line is illustrated in the graph given below. The distance covered by the particle in $4$ seconds is _______m.

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Solution

The correct option is D 55
REF.Image.
distance upto 4 sec
distance $(0 t o 1 s e c) = \frac{1}{2} \times 1 \times 20 = 10 m$
$(1 - 2 s e c) = 20 \times (2 - 1) = 20 m$
$(2 - 3 s e c) = (3 - 2) \times 10 + \frac{1}{2} (3 - 2) \times (20 - 10)$
$= 1 \times 10 + \frac{1}{2} \times 1 \times 10 = 15 m$
$(3 - 4 s e c) = (4 - 3) \times 10 = 10 m$
Total distance = $10 + 20 + 15 + 10 = 55 m$

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The variation of the velocity of a particle moving along a straight line is illustrated in the graph given below. The distance covered by the particle in 4 seconds is _______m.

The correct option is D 55REF.Image.distance upto 4 secdistance (0 to 1sec)=12×1×20=10m(1−2sec)=20×(2−1)=20m(2−3sec)=(3−2)×10+12(3−2)×(20−10)=1×10+12×1×10=15m(3−4sec)=(4−3)×10=10mTotal distance = 10+20+15+10=55m

The variation of the velocity of a particle moving along a straight line is illustrated in the graph given below. The distance covered by the particle in $4$ seconds is _______m.