Answer :
Sure, let's break down the problem step by step to find out how long it took for the airplane to stop.
1. Identify the given values:
- Initial speed of the airplane, [tex]\( u = 51.7 \, \text{m/s} \)[/tex]
- Final speed of the airplane, [tex]\( v = 0 \, \text{m/s} \)[/tex] (since it comes to a stop)
- Acceleration, [tex]\( a = -1.5 \, \text{m/s}^2 \)[/tex] (the negative sign indicates deceleration)
2. Use the kinematic equation:
The equation that relates initial velocity ([tex]\( u \)[/tex]), final velocity ([tex]\( v \)[/tex]), acceleration ([tex]\( a \)[/tex]), and time ([tex]\( t \)[/tex]) is:
[tex]\[
v = u + at
\][/tex]
3. Plug in the known values:
We need to solve for [tex]\( t \)[/tex]:
[tex]\[
0 = 51.7 + (-1.5)t
\][/tex]
4. Solve for [tex]\( t \)[/tex]:
[tex]\[
0 = 51.7 - 1.5t
\][/tex]
[tex]\[
1.5t = 51.7
\][/tex]
[tex]\[
t = \frac{51.7}{1.5}
\][/tex]
[tex]\[
t = 34.47 \, \text{seconds}
\][/tex]
So, it took approximately 34.47 seconds for the airplane to stop.
1. Identify the given values:
- Initial speed of the airplane, [tex]\( u = 51.7 \, \text{m/s} \)[/tex]
- Final speed of the airplane, [tex]\( v = 0 \, \text{m/s} \)[/tex] (since it comes to a stop)
- Acceleration, [tex]\( a = -1.5 \, \text{m/s}^2 \)[/tex] (the negative sign indicates deceleration)
2. Use the kinematic equation:
The equation that relates initial velocity ([tex]\( u \)[/tex]), final velocity ([tex]\( v \)[/tex]), acceleration ([tex]\( a \)[/tex]), and time ([tex]\( t \)[/tex]) is:
[tex]\[
v = u + at
\][/tex]
3. Plug in the known values:
We need to solve for [tex]\( t \)[/tex]:
[tex]\[
0 = 51.7 + (-1.5)t
\][/tex]
4. Solve for [tex]\( t \)[/tex]:
[tex]\[
0 = 51.7 - 1.5t
\][/tex]
[tex]\[
1.5t = 51.7
\][/tex]
[tex]\[
t = \frac{51.7}{1.5}
\][/tex]
[tex]\[
t = 34.47 \, \text{seconds}
\][/tex]
So, it took approximately 34.47 seconds for the airplane to stop.
