Question

A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the

acceleration of the car and the distance traveled.

Answer 1

Thanks for your question !

       The both  accelerations in first and second case are given as follows -
Vi = 18.5 m/s and Vf = 46.1 m/s
and time for both accelerations is 2.47 m/s

       Now, a = v/t
                    = 46.1 - 18.5 / 2.47
                 a = 11.2 m/s^2

                d = vi × t + 0.5 × a × t ^2
                d = 18.5 m/s × 2.47 + 0.5 × (11.2 m/s^2) × (2.47 s)^2
                d = 45.7 m + 34.1 m
                d = 79.8 m

    Therefore, the acceleration of car is 11.2 m/s^2 and distance (travelled) is 79.8 m.

Answer 2

The acceleration of the car and the distance travelled are 11.1 ms-2 and 79.5 m respectively.

Given:

A racecar accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds.

To Find:

The acceleration of the car and the distance travelled.

Solution:

To find the acceleration of the car and the distance travelled we will follow the following steps:

As we know,

The formula for finding acceleration and velocity is

v = u + at

Here, v is the final velocity and u is the initial velocity in ms-1.

a is the acceleration in metres per second square and t is the time in sec.

Now,

Putting values in the above equation we get,

46.1 = 18.5 + 2.47a

46.1 - 18.5 = 2.47a

a = 11.1 ms-2

And,

The formula for finding distance by distance and acceleration is:

$s = ut + \frac{1}{2} a {t}^{2}$

Here, s is the distance in metres.

$s = 45.7 + \frac{1}{2}( 11.1)6 \: = 45.7 \: + 33.8 = 79.5$

The distance travelled Is 79.5 metres.

Henceforth, The acceleration of the car and the distance travelled are 11.1 ms-2 and 79.5 m respectively.