a car moves with a speed of 50km/h. if stopping distance is 40m and its rate of retardation in 4.4m/s^2,then find the time of arrival
Answers
Answered by
0
Answer:50 km/hr=50×
3600
1000
m/s=50×
18
5
=
18
250
m/s
We have, u
2
+2as=0
⇒u
2
=−2as
⇒(
18
250
)
2
=−2a×6
⇒a=−(
18
250
)
2
×
12
1
m/s
2
Now, when speed is:
100 km/hr=100×
18
5
=
18
500
m/s
Again, v
2
=u
2
+2as
⇒0=(
18
500
2
)+2s(−(
18
250
)
2
×
12
1
)
2s=4×12⇒s=24m
Explanation:
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Answered by
0
Answer:
speed of car = 50km/h = 50×5/18
=25×5/9 = 125/9 m/s
stopping distance (s) = 40m
Retardation (a) =-4.4 m/s^2
Time of reaching = ?
from equation motion :
u = u+ at
0 = 125/9-4.4t
t = 125/9×4.4
= 3.16 s
hence, car can be stopped in 3 sec.
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