Question

A car accelerates uniformly from rest and

Answer 1

Answer:

Number of revolutions  = 2.41

Explanation:

According to first equation of motion:

$a=\frac{v-u}{t}$

Here, a= acceleration

v = final velocity

u = initial velocity

According to question ,

a = ?

v = 21.1 m/s

t = 14.1 s

u = 0 (because object starts from rest)

$a=\frac{v-u}{t}$

$a=\frac{21.1-0}{14.1}$

$a=\frac{21.1}{14.1}$

$a=1.49m/s^{2}$

According to third equation of motion:

$2as=v^{2}-u^{2}$

this can also be written as ,

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

s = distance travelled by the object

a = 1.49

$s=\frac{21.1^{2}-0^{2}}{2\times 1.496}$

$s=\frac{21.1^{2}-0^{2}}{2\times 1.496}$

$s=\frac{445.2-0}{2.993}$

$s=\frac{445.2}{2.993}$

s = 148.79 m

Diameter of a tire = 61.7 cm

To calculate number of turns , divide distance by the diameter

$revolutions = \frac{distance}{diameter}$

$= \frac{148.79}{61.7}$

Number of revolutions  = 2.41