A car braked with a constant deceleration of 16 ft/s^2 , producing skid marks measuring 200 ft before coming to a stop. How fast was the car traveling when the brakes were first applied?

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Travis Hicks · Accepted Answer

Step 1 a(t)=−16fts2 We can make it the forward direction is positive, since acceleration is in the opposite direction of forvard if is slowing down. Step 2 Find the antiderivative, which is velocity v(t)=−16t+v0 At this point v0 is the unknown constant, and also what the problem is asking for, the initial speed.t=0 is the time when the brakes are first applied. Step 3 Find the antiderivative, which is the position s(t)=−162t2+v0t+C If we let the position at time t=0 be the 0 position, then C=0 s(t)=−8t2+v0t

Stella Calderon · Accepted Answer

Step 4 When the position is 200 ft, we have s(t)=200=−8t2+v0t 8t2+200=v0t 8t2+200t=v0 Step 5 When the velosity is 0 we have v(t)=0=−16t+v0 16t=v0

karton · Accepted Answer

Step 6 Set the two equal to each other and solve for t. 8t2+200t=16t8t2+200=16t2200=8t2t2=25t=±5 We use the positive number, t=5 s (the time it took to come to a stop) Step 7 Plug into one of the expressions we have for v0 v0=16(5)=80ft/s Result 80 ft/s

A car braked with a constant deceleration of 16 ft/s^2

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