Question

learn more remarks to find the displacement while braking, we could have used the two kinematics equations involving time, namely, $delta x = v_{0}t+\frac{1}{2}at^{2}$ and $v = v_{0}+at$,but because we werent interested in time, the time - independent equation was easier to use. question by how much would the answer change if the plane coasted for 2.0 s before the pilot applied the brakes? enter a number. m

Explanation:

Step1: Analyze the motion during coast - ing

During the 2.0 s of coasting, the plane moves with a constant velocity (assuming no acceleration during coasting). If we assume the initial velocity of the plane before braking is \(v_0\), the displacement during the coast - ing time \(t = 2.0\ s\) can be found using the formula \(x=v_0t\). But since we are not given \(v_0\), we assume the velocity is constant during this non - braking phase. The displacement \(\Delta x_1\) during the 2.0 s of coasting is given by the formula for uniform motion \(\Delta x_1=v_0t\).

Step2: Calculate the additional displacement

We know \(t = 2.0\ s\). Let's assume the velocity of the plane just before braking is \(v_0\). The displacement during the 2.0 s of coasting is \(\Delta x_1\). Using the formula \(x = vt\) (where \(v = v_0\) and \(t=2.0\ s\)), we get \(\Delta x_1=v_0\times2\). If we assume the velocity of the plane before braking is non - zero, the additional displacement due to the 2.0 s of coasting is \(2v_0\). But if we assume the initial velocity \(v_0\) is in m/s, and we consider the motion as a simple uniform motion during coasting, the displacement \(\Delta x\) during the 2.0 s of coasting is calculated as follows:
Let's assume the velocity of the plane before braking is \(v_0\). Using the formula for displacement in uniform motion \(x = vt\), with \(t = 2.0\ s\), we have \(\Delta x=v_0\times2\). If we assume the velocity of the plane before braking is \(v_0 = 10\ m/s\) (for example, since the value of \(v_0\) is not given in the problem, we just show the general method), then \(\Delta x=10\times2=20\ m\). In general, if the velocity of the plane before braking is \(v_0\), the additional displacement due to the 2.0 s of coasting is \(2v_0\). Since we have no information about \(v_0\) other than the motion is uniform during coasting, we assume a general case. The displacement during the 2.0 s of coasting is \(\Delta x = 2v_0\). If we assume the velocity of the plane before braking is non - zero, the answer changes by an amount equal to the displacement during the 2.0 s of coasting. Using the formula \(x = vt\) (uniform motion), with \(t = 2.0\ s\), the additional displacement \(\Delta x\) is given by \(\Delta x=v_0\times2\).

Answer:

We need the initial velocity \(v_0\) of the plane before braking to give a numerical answer. If we assume the initial velocity \(v_0\) is known, the change in the answer (displacement) is \(2v_0\) m. For example, if \(v_0 = 10\ m/s\), the answer is 20 m. But without knowing \(v_0\), we can only say that the change in the displacement is \(2v_0\) m.