Asked by Katelyn
A car is traveling 27 m/s when the driver sees a child standing in the road. He takes 0.8 s to react, then steps on the brakes and slows at 8.0 m/s2. How far does the car go before it stops?
please help!!
Since he's going 27 m/s, and he takes 0.8 s to react, then he goes (27)(0.8) = 21.6 m before he even steps on the break. Then, once he steps on the break, he deceleration is -8.0 m/s^2.
distance = (velocity final)^2 = (velocity initial)^2 + 2ad.
So, velocity final is 0.
So, (velicocity initial)^2 = -2as
(27)^2 = -2(-8.0)d
d = 45.6 m This is the distance after he put on brake.
Total distance = 45.6 + 21.6 m
thank you so much. that answer is right. thank you!!
please help!!
Since he's going 27 m/s, and he takes 0.8 s to react, then he goes (27)(0.8) = 21.6 m before he even steps on the break. Then, once he steps on the break, he deceleration is -8.0 m/s^2.
distance = (velocity final)^2 = (velocity initial)^2 + 2ad.
So, velocity final is 0.
So, (velicocity initial)^2 = -2as
(27)^2 = -2(-8.0)d
d = 45.6 m This is the distance after he put on brake.
Total distance = 45.6 + 21.6 m
thank you so much. that answer is right. thank you!!
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