Question

How long will it take a car to accelerate from 15.2 m/s to 23.5 m/s if the car has an average acceleration of 3.2 m/s²

Answer 1

Answer:

The time taken by the car to change its velocity will be 2.59 seconds (s)

Explanation:

Uniformly Accelerated Motion

We are asked for the time taken to change velocity from initial speed to final speed

Since we know the car initial velocity, the final velocity and the acceleration,

The most suitable motion equation to apply is:

$\large\boxed {\bold {V_{f} = V_{0} \ + a \ . \ t }}$

$\bold { V_{f} } \ \ \ \ \ \ \textsf{ Final Velocity }\ \ \ \bold{23.5\ \frac{m}{s} }$

$\bold { V_{0}} \ \ \ \ \ \ \textsf{ Initial Velocity }\ \ \ \bold{15.2 \ \frac{m}{s} }$

$\bold { a }\ \ \ \ \ \ \ \ \textsf{ Acceleration }\ \ \ \bold{3.2 \ \frac{m}{s^{2} } }$

$\bold { t} \ \ \ \ \ \ \ \ \textsf{ time taken to change velocity }\ \ \ \bold{ s }$

Thus we have

$\boxed {\bold {V_{f} = V_{0} \ + a \ . \ t }}$

$\boxed {\bold {V_{f} - V_{0} \ = a \ . \ t }}$

$\large\boxed {\bold { t = \frac{V_{f} \ -\ V_{0} }{ a\ } }}$

$\large\textsf{ Substituting the given values and solving }$

$\boxed {\bold { t = \frac{23.5 \ \frac{m}{s} \ -\ 15.2 \ \frac{m}{s} }{ 3.2 \ \frac{m}{s^{2} } } }}$

$\boxed {\bold { t = \frac{8.3 \ \frac{\not m}{\not s} }{ 3.2 \ \frac{\not m}{s^{\not 2} } } }}$

$\large\boxed {\bold { t = 2.59 \ s }}$

Hence the time taken to change velocity will be 2.59 seconds (s)