Question
A car traveling in a straight ine at the initial speed of 8 m/s accels by 14 uniformly over a distance of 44 meters what's the mag of aceel of the car
Answers
GPT-4o mini
To find the magnitude of the acceleration of the car, we can use the kinematic equation which relates initial velocity, final velocity, acceleration, and displacement:
\[
v^2 = u^2 + 2as
\]
Where:
- \( v \) = final velocity (m/s)
- \( u \) = initial velocity (8 m/s)
- \( a \) = acceleration (m/s²)
- \( s \) = displacement (44 m)
We need to rearrange this equation to solve for \( a \):
\[
a = \frac{v^2 - u^2}{2s}
\]
We are not given the final velocity directly, but we know that the car accelerates uniformly over the distance. The acceleration is given as 14 m/s² (likely to clarify the prompt about the acceleration).
Assuming the problem intended to say the car has an acceleration of 14 m/s² and you need to check if this results in an appropriate final velocity over 44 m:
With \( u = 8 \, \text{m/s} \), \( a = 14 \, \text{m/s}^2 \), and \( s = 44 \, \text{m} \):
\[
v^2 = u^2 + 2as
\]
\[
v^2 = (8)^2 + 2(14)(44)
\]
\[
v^2 = 64 + 1232
\]
\[
v^2 = 1296
\]
\[
v = \sqrt{1296} = 36 \, \text{m/s}
\]
Thus, if the acceleration is indeed 14 m/s², then the car would reach a final speed of 36 m/s over the 44 m distance.
So the magnitude of the acceleration of the car is \( 14 \, \text{m/s}^2 \).
\[
v^2 = u^2 + 2as
\]
Where:
- \( v \) = final velocity (m/s)
- \( u \) = initial velocity (8 m/s)
- \( a \) = acceleration (m/s²)
- \( s \) = displacement (44 m)
We need to rearrange this equation to solve for \( a \):
\[
a = \frac{v^2 - u^2}{2s}
\]
We are not given the final velocity directly, but we know that the car accelerates uniformly over the distance. The acceleration is given as 14 m/s² (likely to clarify the prompt about the acceleration).
Assuming the problem intended to say the car has an acceleration of 14 m/s² and you need to check if this results in an appropriate final velocity over 44 m:
With \( u = 8 \, \text{m/s} \), \( a = 14 \, \text{m/s}^2 \), and \( s = 44 \, \text{m} \):
\[
v^2 = u^2 + 2as
\]
\[
v^2 = (8)^2 + 2(14)(44)
\]
\[
v^2 = 64 + 1232
\]
\[
v^2 = 1296
\]
\[
v = \sqrt{1296} = 36 \, \text{m/s}
\]
Thus, if the acceleration is indeed 14 m/s², then the car would reach a final speed of 36 m/s over the 44 m distance.
So the magnitude of the acceleration of the car is \( 14 \, \text{m/s}^2 \).