Question

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Answer 1

1)ABCD is a square field of side 10 m
The farmer moves along the boundary of the field from the corner A via the corners B, C and D. After every 40 s, the farmer is again at the corner A, so his displacement after every 40s is zero. Let us convert 2mins 20 s=(2x60+20)s =140 s, the farmer will be at the corner C.
therefore,the magnitude of his displacement =>AC= √Ab²+√BC²
=>AC=√10²+√10²=√200=10√2 m =14.4 m

2)Here
i),u=5m/s
a= -10m/s²
v=0
s=?
t=? we get,
s=v²-u²/2a
s= 0-25/-2 x10
s=1.25 m
II) using v=u+at we get,
=>t=v-u/a
=>t=0-5/-10
=>t=0.5 s

3)Here,a)
u=0
a=0.2 m/s²
t=2 mins =120s
using v=u+at we get,
v=0+0.1 x 120 =12m/s
speed acquired =12m/s
b) using, s=ut+1/2at²
s=0+1/2 x0.1 x120 x 120
s=720 m.
distance travelled =720 m.

4)when an object moves in one direction along a straight line.

5)Here u=80km/h =80x 5/18 m/s
=22.22 m/s
v=60km/h =60 x5/18 m/s
=16.67 m/s
t=5s
therefore a=v-u/t
a=16.67-22.22/5
a= -5.55/5
a= -1.11m/s²

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Answer 2

5)given- v = 80km / h = 80*1000/60*60=22.2 m/s. t = 5 sec.
u =60 km / h = 60*1000/60*60=1.6 m/s.
Now, putting formula of accerelation =
a = v - u / t = 22.2 -1.6 / 5 = 20.6 / 5 = 4.12 m/s