Answer:
[tex]\boxed {\boxed {\sf 253 \ meters}}[/tex]
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:
[tex]d=v_it+\frac{1}{2}at^2[/tex]
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.
- [tex]v_i[/tex]= 0 m/s
- a= 0.250 m/s²
- t=45.0 s
Substitute the values into the formula.
[tex]d= (0 \ m/s)(45.0 \ s) + \frac{1}{2} (0.250 \ m/s^2)(45.0 \ s)^2[/tex]
Multiply the first two numbers. The units of seconds cancel.
[tex]d= 0 \ m + \frac{1}{2} (0.250 \ m/s^2)(45.0 \ s)^2[/tex]
Solve the exponent.
- (45.0 s)² = 45.0 s *45.0 s = 2025 s²
[tex]d= 0 \ m +\frac{1}{2} (0.250 \ m/s^2)(2025 \ s^2)[/tex]
Multiply the numbers in parentheses. The units of seconds squared cancel.
[tex]d= 0 \ m +\frac{1}{2} (0.250 \ m/s^2 * 2025 \ s^2)[/tex]
[tex]d= 0 \ m +\frac{1}{2} (0.250 \ m * 2025 )[/tex]
[tex]d= 0 \ m + \frac{1}{2} (506.25 \ m)[/tex]
Multiply by 1/2 or divide by 2.
[tex]d=0 \ m + 253.125 \ m \\d= 253.125 \ m[/tex]
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.
[tex]d \approx 253 \ m[/tex]
The train travel approximately 253 meters.
