Question

A car traveling in a straight line at an initial speed of 8.0 meters per second accelerates uniformly to a speed of 14 meters per second over a distance of 44 meters. What is the magnitude of the acceleration of the car?

Answer 1

The equation we can use here is:

v^2 = v0^2 + 2 a d

where v is final velocity, v0 is intial velocity, a is acceleration and d is distance

 

14^2 = 8^2 + 2 a (44)

a = 1.5 m/s^2

Answer 2

The acceleration of the car travelling in a straight line is $\fbox{\begin\\1.5\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\end{minispace}}$.

Further Explanation:

A car travelling in a straight line with a constant uniform acceleration will always follow the Newton’s law of motion.

Given:

The initial velocity of the car is $8\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$.

The final velocity of the car is $14\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$.

The total distance covered by the car is $44\,{\text{m}}$.

Concept:

The car is following the Newton’s law of motion. As per the Newton's laws of motion, the motion of a body is governed by the three equations of motion.

To calculate the uniform acceleration of car when initial velocity, final velocity, and distance covered by car given, the equation of motion we can use here is:

$\fbox{\begin\\v^2 = {u^2} + 2as\end{minispace}}$

Rearrange the above equation for value of acceleration :

$a=\dfrac{{{v^2}-{u^2}}}{{2s}}$

Here, $s$ is the distance covered by car, $u$ is the given initial velocity of the car, $v$ is the given final velocity of the car and $a$ is the acceleration of the car.

Substitute $14\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for $v$, $8\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for $u$ and $44\,{\text{m}}$ for $s$ in the above equation.

$\begin{aligned}a&=\dfrac{{{{\left({14\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}\right)}^2}-{{\left({8\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}\right)}^2}}}{{2\times44}}\\&=1.5\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{aligned}$

Thus, the acceleration of the car travelling in a straight line is $\fbox{\begin\\1.5\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\end{minispace}}$.

Learn more:

1.  Conservation of momentum brainly.com/question/9484203

2.  Motion of a ball under the gravity brainly.com/question/10934170

3. The motion of a body under the friction force brainly.com/question/4033012

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

Initial speed of 8.0 m/s,Newton's law, motion, distance, final speed of 14.0 m/s, car travelling in a straight line, accelerates uniformly and distance of 44 m.