\\PROBLEM-SOLVING STRATEGY: Motion with constant acceleration Problems involving

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PROBLEM-SOLVING STRATEGY: Motion with constant acceleration

Problems involving constant accelerationspeeding up, slowing down, vertical motion, horizontal motioncan all be treated with the same problem-solving strategy.

PREPARE: Draw a visual overview of the problem. This should include a motion diagram, a pictorial representation, and a list of values; a graphical representation may be useful for certain problems.

SOLVE: The mathematical solution is based on the three equations

(vx)f=(vx)i+axt,

xf=xi+(vx)it+12ax(t)2,

(vx)2f=(vx)2i+2axx.

Though the equations are phrased in terms of the variable x, it is customary to use y for motion in the vertical direction.

Use the equation that best matches what you know and what you need to find. For example, if you know acceleration and time and are looking for a change in velocity, the first equation is the best one to use.

Uniform motion with constant velocity has zero acceleration (a=0).

ASSESS: Is your result believable? Does it have proper units? Does it make sense?

Part B

Sort the variables for the period from t1=0s to t2=0.5s based on whether their values are known or unknown.

Part C

Sort the variables for the period from t2=0.5s until the car comes to rest, at time t3, based on whether their values are known or unknown.

Explanation / Answer

Part B

for t1=0s to t2=0.5s and acceleration ax =0

(vx)f=(vx)i+0(0.5-0)

(vx)f=(vx)i

xf=xi+(vx)i*0.5+1/2* 0*(0.5)2

xf=xi+(vx)i*0.5

(vx)2f=(vx)2i+2*0*0.5

(vx)2f=(vx)2i

Part C: for time 0.5sec to comes to rest

(vx)f=(vx)i+0xt,

xf=xi+(vx)i

xf=xi+(vx)it+12*0*t)2

xf=xi+(vx)i*1.5

(vx)2f=(vx)2i+2*0x

(vx)2f=(vx)2i

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