First slide

The 3 equation for uniform acc

Question

The displacement of a particle moving in a straight line is described by the relation $S = 6 + 12 t - 2 t^{2}$ . Here S in meter and t is in sec. The distance covered by the particle in first 5s is

Difficult

A

20 m
B

24 m
C

16 m
D

26 m

Solution

s=6+12t−2t²
differentiating with respect to t
v=12−4t
direction is reversed , v=0
at time, t = 3s
At t=0,
The displacement is 6 meters
displacement in 3 secs is 6+12(3)−2(3)²=24

displacement in 5 secs is 6+12(5)−2(5)²=16

Distance travelled in forward direction in 3 sec is 24-6=18 m
Distance travelled in reverse direction in next 2 sec is 24-16=8 m
Thus total distance travelled in 5 sec= 18+8=26 m

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