Video: Understanding What Is Meant by Constant Acceleration

A sphere is released from rest from the top of a 50-meter-long inclined plane. One second after the sphere is released it has rolled 3 meters along the plane. Given that the sphere has a constant acceleration, how far will it roll along the plane in 4 seconds? [A] 24 m [B] 48 m [C] 12 m [D] 36 m [E] 50 m

05:03

Video Transcript

A sphere is released from rest from the top of a 50-meter-long inclined plane. One second after the sphere is released, it has rolled three meters along the plane. Given that the sphere has a constant acceleration, how far will it roll along the plane in four seconds? a) 24 meters, b) 48 meters, c) 12 meters, d) 36 meters, e) 50 meters.

Since this is a mechanics question about the motion of an object, drawing a labeled picture will help us organize our thoughts. Here, we’ve drawn the inclined plane and the sphere. Now, let’s label what we know. The inclined plane is 50 meters long. So we’ve leveled the top of the inclined plane zero meters and the bottom of the inclined plane 50 meters. And we’ll use the coordinate 𝑥, which increases as we move down the plane, to represent our distance in meters from the top of the plane.

Let’s now write down what we know about the sphere. 𝑥 zero, the initial position of the sphere, is zero meters, since we’re told that the sphere starts at the top of the plane and we’ve defined this position to be zero. 𝑣 zero, the initial velocity of the sphere, is also zero, since we’re told that the sphere is released from rest and, by definition, an object at rest has zero velocity.

We’re also told some information about the sphere at time 𝑡 equals one second after release. One second after the sphere is released, it has rolled three meters along the plane. So the 𝑥-coordinate of the sphere, its position along the inclined plane, evaluated at one second is three meters. The last piece of information that we’re given is that the sphere has a constant acceleration throughout its motion. We’ll use the symbol 𝑎 for acceleration and we’ll write 𝑎 equals constant.

Finally, the essence of the question itself is how far will the sphere roll along the plane in four seconds. Using the symbols we’ve defined, we’re looking for the 𝑥-coordinate of the sphere at four seconds. What we need now is some equation that we can use to combine our given quantities into the quantity that we’re looking for. Since the sphere is moving in only one dimension, 𝑥, and since the acceleration is constant and we have information about the initial position and initial velocity, this suggests that the following kinematic equation is applicable. The position at the time of interest 𝑡 is equal to the initial position plus the initial velocity times the time plus one-half of the acceleration times time squared.

This equation is valid under the assumption that the motion is one dimensional and that the acceleration is constant, exactly what we have in our problem. The sphere is moving only in the dimension 𝑥 and its acceleration is constant. Let’s plug in values to find the position of the box at four seconds. Using zero meters as the initial position and zero meters per second as the initial velocity, we see that the position of the sphere at four seconds is zero meters plus zero meters per second times four seconds plus one-half 𝑎 times four seconds squared.

But zero meters per second times four seconds is just zero. And zero meters is also zero. So the only nonzero portion of this sum is one-half 𝑎 times four seconds squared. So the position of the sphere at four seconds is one-half 𝑎 times four seconds squared. Four seconds squared is 16 seconds squared divided by two is eight seconds squared. The trouble is we don’t know the value of acceleration, only that it’s constant. We do, however, know the position of the sphere at a different time. At one second, the position of the sphere is three meters. So let’s write down our kinematic equation for 𝑡 equals one second.

Three meters, the position of the sphere at time 𝑡 equals one second, is zero meters, the initial position, plus zero meters per second times one second, the initial velocity times the time, plus one-half 𝑎 times one second squared. But just like before, zero meters and zero meters per second times one second are both zero. And we’re left with three meters is equal to one-half 𝑎 times one second squared. One-half times one second squared is just one-half second squared. Now, let’s multiply both sides by two and divide by second squared.

On the right-hand side, second squared divided by second squared is one. And two times one-half is also one. So the right-hand side is just equal to 𝑎. On the left-hand side, two times three meters is six meters divided by seconds squared is per second squared. So the constant acceleration of the sphere is six meters per second squared. Let’s substitute this value for the acceleration of the sphere into our expression for the position of the sphere at 𝑡 equals four seconds.

The position of the sphere along the inclined plane at 𝑡 equals four seconds — that is, how far it’s rolled — is equal to six meters per second squared times eight seconds squared. Seconds squared divided by seconds squared is just one. And we’re left with eight times six meters, which is equal to 48 meters. So after four seconds, the sphere has rolled 48 meters down the inclined plane, and the correct answer choice is b.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.