Calculating average acceleration on an inclined plane

Key concept: This kinematics question uses the displacement-time relation for constant average acceleration. Start by finding the acceleration from motion over the given interval, then interpret it in the direction of the incline.

Step by step

Key idea: use kinematics over the time interval

The problem gives a displacement of 4.00 m in 4.00 s down the incline. Since it asks for average acceleration, the simplest approach is to use the motion relation for constant acceleration from rest or to interpret the average change in velocity over time if the motion is assumed uniform over the interval.

For introductory incline questions like this, the standard calculation usually uses

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\$$ when the object starts from rest. Solving for acceleration gives \ $$a = \frac{2s}{t^2} \$$ ## Step 1: substitute the given values \ $$a = \frac{2(4.00)}{(4.00)^2} \$$ \ $$a = \frac{8.00}{16.0} = 0.500\,\text{m/s}^2 \$$ ## Step 2: state the direction Because the book slides down the incline, the acceleration is directed down the slope. ## Step 3: interpret the result An acceleration of \$0.500\\,\text{m/s}^2\$ is relatively small, which suggests that friction may be reducing the component of gravity along the incline. The incline angle alone would not be enough to determine friction, but the measured motion tells us the net acceleration. ## Final answer \ $$\boxed{0.500\,\text{m/s}^2\text{ down the incline}}\$$ ### Pitfall alert One common mistake is treating the 4.00 m as a velocity or using \$a=s/t\$, which is not a valid acceleration formula. Another error is ignoring direction and giving only a number, even though motion down the incline has a specific direction. If the object is not stated to start from rest, you should be careful with formulas that assume zero initial velocity; in that case, the question may require a different kinematic relation or more information. Always check whether the wording implies rest, constant acceleration, or only an average over the interval. ### Try different conditions If the same setup said the textbook started with an initial velocity of 1.00 m/s down the incline, then you could not use \$s=\tfrac12 at^2\$ alone. You would need \$s = v_0 t + \tfrac12 at^2\$, which would change the acceleration value. If the distance were 6.00 m in 4.00 s instead, and the object still started from rest, the acceleration would become \$a=\frac{2(6.00)}{16.0}=0.750\\,\text{m/s}^2\$. ### Further reading kinematic equation, average acceleration, inclined plane motion