1.A car is traveling at 16 ft/s when the brakes are fully applied, producing a d

Question

1.A car is traveling at 16 ft/s when the brakes are fully applied, producing a deceleration a(t) = 5 t2/3 ft/s2. What is the distance covered before the car comes to a stop? 1.A car is traveling at 16 ft/s when the brakes are fully applied, producing a deceleration a(t) = 5 t2/3 ft/s2. What is the distance covered before the car comes to a stop? 1.A car is traveling at 16 ft/s when the brakes are fully applied, producing a deceleration a(t) = 5 t2/3 ft/s2. What is the distance covered before the car comes to a stop?

Explanation / Answer

u=16 ft/s

v=0

a(t)=5 t^(2)/3 ft/s^2

v=u+at

0=16(5 t^(2)/3)*t

(5 t^(2)/3)*t=16

t^3=48/5

t=(48/5)^(1/3)

s=ut+0.5at^2

s=16*(48/5)^(1/3)-0.5*16*(48/5)^(-1/3)*(48/5)^(2/3)=(48/5)^(1/3)[16-8]=8*(48/5)^(1/3) ft

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