HI I AM KIND OF STUCK ON P 27 EDITION 4TH OF CUTNELL &JOHNSONBOOK. I KEEP GETTIN

Question

HI

I AM KIND OF STUCK ON P 27 EDITION 4TH OF CUTNELL &JOHNSONBOOK. I KEEP GETTING - 2.94M/S2 FOR THE ACCELERATION WHILE THE BOOKANSWER IS -4.6. I STILL DO NOT SEE HOW THEY GET THERE.

data: a speed trap is set up with two pressure-activatedstrips placed across a highway, 110 m apart. A car speeding alongat 33m/s, while the speed limit is 21m/s.At the instant the caractivates the first strip, the driver begins slowing down. Whatminimum deceleration is needed in order that the average speed iswithin the speed limit by the time the car crosses the secondmarker.?

I used the equation v2- vo2= 2ax with V= 21m/s Vo= 33m/s andX= 110 m and my answer is - 2.9m/s2

Please help

Forna Diphicyl HI

I AM KIND OF STUCK ON P 27 EDITION 4TH OF CUTNELL &JOHNSONBOOK. I KEEP GETTING - 2.94M/S2 FOR THE ACCELERATION WHILE THE BOOKANSWER IS -4.6. I STILL DO NOT SEE HOW THEY GET THERE.

data: a speed trap is set up with two pressure-activatedstrips placed across a highway, 110 m apart. A car speeding alongat 33m/s, while the speed limit is 21m/s.At the instant the caractivates the first strip, the driver begins slowing down. Whatminimum deceleration is needed in order that the average speed iswithin the speed limit by the time the car crosses the secondmarker.?

I used the equation v2- vo2= 2ax with V= 21m/s Vo= 33m/s andX= 110 m and my answer is - 2.9m/s2

Please help

Forna Diphicyl

Explanation / Answer

we need the average velocity to be 21m/s so that our friend doesn'tget a ticket so we know the distance between the two strips and wecan find the minimum amount of time it should take to make thetrip so v = d/t or t = d/v which means it should take him 5.238seconds we are going to use the kinematic equation x = xinit +vinit t + 1/2 a t2 so we now have: 110 = 0 + 33m/s (5.23s) + 1/2 a (5.23)2 -62.857 = 1/2 a 27.35 and we find that a = -4.596m/s2

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