A. 15 m
B. 20 m
C. 5 m
D. 25 m
answer is 15 m
S=ut+1/2gt^2
u=0 ,g = 10ms-2
During ist time interval
S1=0+1/2(10)(1)^2
S1=5m
During 2nd time interval
S2=0+1/2(10)(2)^2
S2=20m
Distance covered by body during an interval of time between first and second seconds of its motion = (S2 – S1 ) = 20–5 = 15m .
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PLEASE COMMENT BELOW WITH CORRECT ANSWER AND ITS DETAIL EXPLANATION.
Using the kinematics equation, s = u.t + 1/2.a.t².
In t = 1sec, s = 1/2 (10) (1)² = 5m
In t = 2sec, s = 1/2 (10) (2)² = 20m
The distance travelled between the first and second seconds = 20 – 5 = 15m
It is wrong answer the correct is 15m
how
It will be 5.
Using second equation of motion
S=vit+1/2gt2
So here initial velocity is zero and time between two interval is t2-t1
So t =2-1=1
So result will b 5m
How this answer is 14?which is suitable formula for this question
how can we explain it?????????plz
Hello this question answer is 15m
S1= 1/2gt^2 = 1/2(10)(1)^2 = 5m and
S2 = 1/2gt^2 = 1/2(10)(2)^2 = 1/2(10)(4) = 20m
so S = S2-S1
S =20 – 5 = 15m
5m
D1=VT +1/2gt². At 2 second
D1= 0+1/2*10*4
D1= 40/2 =20m
At one second
D2=VT +1/2gt².
D2=0+1/2*10*1
D2= 5 m
D=D2-D1
D=20-5 =15 m
answer is 15 meter
answer is 15 m
S=ut+1/2gt^2
u=0 ,g = 10ms-2
During ist time interval
S1=0+1/2(10)(1)^2
S1=5m
During 2nd time interval
S2=0+1/2(10)(2)^2
S2=20m
Distance covered by body during an interval of time between first and second seconds of its motion = (S2 – S1 ) = 20–5 = 15m . Hope you like it!
could someone plz solve this question
Explanation would have been helpful….
Answer is 20
The answer is 20 m, kindly correct this