a car is traveling at 20 m per second along a road a child run out into the road 50m ahead and the car driver step on the brake pedal what must the cars deceleration be if the car is to stop just before it reach the child
Answers
Answered by
2
at deceleration a, the velocity is
v = 20-at
so t = 20/a
the distance traveled by the car is
s = 20t + 1/2 at^2
= 20(20/a) - 1/2 a (20/a)^2
= 200/a
So, to stop in 50 meters,
200/a = 50
a = 4 m/s^2
check:
v = 20-4*5
s = 20*5 - 2*5^2 = 100-50 = 50
v = 20-at
so t = 20/a
the distance traveled by the car is
s = 20t + 1/2 at^2
= 20(20/a) - 1/2 a (20/a)^2
= 200/a
So, to stop in 50 meters,
200/a = 50
a = 4 m/s^2
check:
v = 20-4*5
s = 20*5 - 2*5^2 = 100-50 = 50
Answered by
2
u= 20 m/s
s= 50 m
v=0
Now, With the help of Newton's 3rd equation,
v^2 = u^2 + 2as
0 = (20)^2 + 2a× 50
100a = -400
a = -400/100
a= -4 m/s^2
So, deceleration (a) = -4 m/s^2
s= 50 m
v=0
Now, With the help of Newton's 3rd equation,
v^2 = u^2 + 2as
0 = (20)^2 + 2a× 50
100a = -400
a = -400/100
a= -4 m/s^2
So, deceleration (a) = -4 m/s^2
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