To find the acceleration and distance traveled by the car, we can use the equations of motion.
Given: Initial velocity (uuu) = 0 kph (since the car starts from rest) Final velocity (vvv) = 15 kph Time (ttt) = 2 seconds
First, let's convert the velocities to meters per second (m/s) since the equations of motion require the use of SI units:
Initial velocity (uuu) = 0 kph = 0 m/s Final velocity (vvv) = 15 kph = 15 * (1000/3600) m/s ≈ 4.17 m/s Time (ttt) = 2 seconds
The first equation of motion relates acceleration (aaa), initial velocity (uuu), final velocity (vvv), and time (ttt): v=u+atv = u + atv=u+at
Substituting the known values: 4.17 m/s = 0 m/s + a * 2 s
Simplifying the equation: 4.17 m/s = 2a
Solving for acceleration (aaa): a=4.17 m/s2 sa = frac{4.17 , ext{m/s}}{2 , ext{s}}a=2s4.17m/s a≈2.085 m/s2a approx 2.085 , ext{m/s}^2a≈2.085m/s2
The acceleration of the car is approximately 2.085 m/s22.085 , ext{m/s}^22.085m/s2.
To find the distance traveled (ddd), we can use the second equation of motion: d=ut+12at2d = ut + frac{1}{2}at^2<span class="strut"