Question

A train accelerates from 30 km/h to 45 km/h in 15.0 second. Find its acceleration and the distance it travels during this time

Answer 1

Answer:

a. Acceleration, a = 0.28 m/s²

b. Distance, S = 156 meters

Explanation:

Given the following data;

Initial velocity = 30 km/h

Final velocity = 45 km/h

Time = 15 seconds

a. To find the acceleration;

Conversion:

30 km/h to m/s = 30*1000/3600 = 8.33 m/s

45 km/h to m/s = 45*1000/3600 = 12.5 m/s

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation;

$a = \frac{12.5 - 8.3}{15}$

$a = \frac{4.2}{15}$

Acceleration, a = 0.28 m/s²

b. To find the distance travelled, we would use the second equation of motion given by the formula;

$S = ut + \frac {1}{2}at^{2}$

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

$S = 8.3*15 + \frac {1}{2}*(0.28)*15^{2}$

$S = 124.5 + 0.14*225$

$S = 124.5 + 31.5$

S = 156 meters