Help with physics acceleration problem?

Question

This problem has me stumped...if someone could get the process started for me:

A car is on a dry road with a velocity of 32 m/s. The driver brakes and the car halts with an acceleration of -8.00 m/s^2. On an icy road, the acceleration would have been -3.00 m/s^2. How much further would the car have skidded on the icy road?

Answer 1

I teach physics. Email me anytime with physics questions
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Probably the easiest way for a beginner to do this is to solve the problem twice. Once for the dry conditions, and once for the wet conditions. Then compare the two answers you get to determine the difference between the two answers.

Each of the two times you solve it, you have two problems to work (that makes 4 problems total plus a subtraction)

First, use the speed and acceleration to determine the time it's skidding. The equation is
v = a t

Second, use the time you found to determine how far it skidded using
x = 1/2 (a) (t)^2

First half is done. Repeat those two steps for the icy road, then compare the answers.

Feel free to email me with what you think is the correct question/answer anytime
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- Kevin

Answer 2

Imagine the question in reverse: from a standstill, what distance will the car cover, at the given accelerations, in getting up to 32 m/s?

Answer 3

If the boy drops the ball to loose fall, the next will have to practice Initial Velocity is zero m/s Final Velocity is 14.7 m/s Acceleration is (gravity =nine.eight m/s2) distance = one million/two * gt^two = one million/two * nine.eight * four = 19.6 m If the boy vigour throws the ball with preliminary velocity of 14.7 m/s, then the next will have to practice d = Vi *t + one million/two gt^two d= 14.7*two + one million/two*nine.eight*four = forty nine m

Answer 4

I'm also a physics teacher. Kevin gave you the best approach.

Good luck.