a car moving with a velocity of 10m/s covers a distance of 200 m. Calculate the time taken.
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Case I. Suppose reduction time = t , so = 10 meters
then distance in t sec will be 10t meter
⇒ remaining distance = 10 meter - 10 t
using v
2
=u
2
−2as
we get s=
2a
u
2
−v
2
=
2a
(10)
2
−0
=
a
50
⇒
a
50
=10−10t _______ (1)
Case II
u=20 m/s & S
0
=30 meter
distance in t sec will be 20 t meter
distance during retardation S=
2a
u
2
−v
2
=
2a
(20)
2
−0
S=
a
200
S
0
=30m⇒
a
200
+20t=30
a
50
+5t=7.5 _______ (2)
using (1)
a
50
+10t=10
- - -
___________________
0-5t= -2.5 ⇒ subtracting t=0.5 second
solution
Case I. Suppose reduction time = t , so = 10 meters
then distance in t sec will be 10t meter
⇒ remaining distance = 10 meter - 10 t
using v
2
=u
2
−2as
we get s=
2a
u
2
−v
2
=
2a
(10)
2
−0
=
a
50
⇒
a
50
=10−10t _______ (1)
Case II
u=20 m/s & S
0
=30 meter
distance in t sec will be 20 t meter
distance during retardation S=
2a
u
2
−v
2
=
2a
(20)
2
−0
S=
a
200
S
0
=30m⇒
a
200
+20t=30
a
50
+5t=7.5 _______ (2)
using (1)
a
50
+10t=10
- - -
___________________
0-5t= -2.5 ⇒ subtracting t=0.5 second
solution
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