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 sand 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.

Short answer

the total distance covered is 1796.48 m.

Step-by-step solution

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 $3.50\frac{m}{s^2}$.

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.

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-

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

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:

role="math" localid="1668403990859" $\begin{array}{l}s=vt\\ =22.4\frac{m}{s}\times 70s\\ =1568m\end{array}$

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

role="math" localid="1668404034534" $v^2=u^2+2as$

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-

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

Thus, the total distance covered is 1796.48 m.