Question

A train leaving the station accelerates from a full stop at a rate of 0.250 meters per second squared for 45.0 seconds. How far does the train travel during this time?​

Answer 1

Answer:

$\boxed {\boxed {\sf 253 \ meters}}$

Explanation:

We are asked to find the distance a train travels.

We are given the acceleration, initial velocity, and time. Therefore, we will use the following kinematic equation:

$d=v_it+\frac{1}{2}at^2$

The train starts at a full stop or a velocity of 0 meters per second. It accelerates at a rate of 0.250 meters per second squared for 45.0 seconds.

• $v_i$= 0 m/s
• a= 0.250 m/s²
• t=45.0 s

Substitute the values into the formula.

$d= (0 \ m/s)(45.0 \ s) + \frac{1}{2} (0.250 \ m/s^2)(45.0 \ s)^2$

Multiply the first two numbers. The units of seconds cancel.

$d= 0 \ m + \frac{1}{2} (0.250 \ m/s^2)(45.0 \ s)^2$

Solve the exponent.

• (45.0 s)² = 45.0 s *45.0 s = 2025 s²

$d= 0 \ m +\frac{1}{2} (0.250 \ m/s^2)(2025 \ s^2)$

Multiply the numbers in parentheses. The units of seconds squared cancel.

$d= 0 \ m +\frac{1}{2} (0.250 \ m/s^2 * 2025 \ s^2)$

$d= 0 \ m +\frac{1}{2} (0.250 \ m * 2025 )$

$d= 0 \ m + \frac{1}{2} (506.25 \ m)$

Multiply by 1/2 or divide by 2.

$d=0 \ m + 253.125 \ m \\d= 253.125 \ m$

The original measurements of acceleration and time both have 3 significant figures, so our answer must have the same. For the number we found, that is the ones place. The 1 in the tenths place tells us to leave the 3.

$d \approx 253 \ m$

The train travel approximately 253 meters.