Question

198.
The variation of velocity of a particle moving along a
straight line is illustrated in the following figure.
203
A
B
20+
C
u(ms)
10+
204
0
1
2
3
4
t(s)
The distance traversed by the particle in 4 seconds is
(A) 60 m
(B) 25 m
(C) 55 m
(D) 30 m
(Kerala, 1982)
20​

Answer 1

Answer:

distance upto 4 sec

distance (0 to 1sec)=21×1×20=10m

(1−2sec)=20×(2−1)=20m

(2−3sec)=(3−2)×10+21(3−2)×(20−10)

=1×10+21×1×10=15m

(3−4sec)=(4−3)×10=10m

Total distance = 10+20+15+10=55m