Q60P - University Physics with Modern Physics

Question

A subway train starts from rest at a station and accelerates at a rate of $\mathbf{160}\frac{\mathbf{m}}{{\mathbf{s}}^{\mathbf{2}}}$ for 14s. It runs at constant speed for 70.0 s and slows down at a rate of $\mathbf{3}\mathbf{.}\mathbf{50}\frac{\mathbf{m}}{{\mathbf{s}}^{\mathbf{2}}}$ until it stops at the next station. Find the total distance covered.

Step-by-Step Solution

Answer

the total distance covered is 1796.48 m.

Step 1: Identification of the given data

The given data can be expressed below as:

• The train accelerates at a rate of $1.60\frac{m}{s^2}$.
• The train accelerates in 14 s.
• The train runs at a constant speed of 70.0 s.
• The train slows down at a rate of $3.50\frac{m}{s^2}$.

Step 2: Significance of the Newton’s first law for the subway train

This law signifies that a particular object will continue to move at a constant velocity unless the object is gets resisted by an external force.

The equation of the velocity of the object gives the distance covered by the object.

Step 3: Determination of the total distance covered

From Newton’s first law, the velocity of the subway train for the first 14 s is expressed as:

v = u + at

Here, v is the final velocity of the subway train, u is the initial velocity of the train which is 0 as the train was at rest. The value of the acceleration and the time taken by the subway train are $160\frac{m}{s^2}$ and 14 s respectively.

Substituting the values in the above expression, we get-

$\begin{array}{l}v=0+1.60 m/s^2\times 14s\\ V=22.4\frac{m}{s}\end{array}$

Hence, the distance of the subway train is:

$\begin{array}{r}s=ut+\frac{1}{2}a{t}^2 \\ s=0\times 14\mathrm{s}+\frac{1}{2}\times 1.60\mathrm{m}/{\mathrm{s}}^2\times (14\mathrm{s}{)}^2\\ =156.8\mathrm{m} \end{array}$

Hence, for the next 70.0 s, the displacement of the subway train is expressed as:

$\begin{array}{l}s=vt\\ =22.4\frac{m}{s}\times 70s\\ =1568m\end{array}$

As the subway train slows down at a rate of $3.50 m/s^2$, the distance covered by the train after is expressed as:

$v^2=u^2+2as$

Here, v is the final velocity of the train which is 0 as the train comes to rest, Hence, the initial velocity u becomes the final velocity which is 22.4 m/s, the acceleration of the train is $-3.50m/s^2$ as it slows down.

Substituting the values in the above equation, we get-

$\begin{array}{r}0=(22.4\mathrm{m}/\mathrm{s}{)}^2+2\times \left(-3.5\mathrm{m}/{\mathrm{s}}^2\right)\times s\\ -501.76{\mathrm{m}}^2/{\mathrm{s}}^2=-7\mathrm{sm}/{\mathrm{s}}^2 \\ s=71.68\mathrm{m} \end{array}$

So, the total distance travelled by the subway train is-

$\begin{array}{l}s=71.68m+1568 m+156.8 m\\ s=1796.48m\end{array}$ 

Thus, the total distance covered is 1796.48 m.