Question

1 3.3.A car is travelling with velocity 18km/hr changes to 72km/hr in 10 sec. Find the distance travelled by the body​

Answer 1

Answer:

125 m

Explanation:

As per the provided information in the given question, we have:

• Initial velocity (u) = 18 km/h
• Final velocity (v) = 72 km/h
• Time taken (t) = 10 seconds

We are asked to calculate the distance travelled by the body.

In order to calculate the distance travelled by the body, firstly we need to find the acceleration of the body. In order to do that so, we need to convert velocities in m/s.

Converting initial velocity in m/s :

$\longmapsto\rm{u = 18 \; kmh^{-1} }$

$\longmapsto\rm{u =\Bigg ( 18 \times \dfrac{5}{18} \Bigg ) \; ms^{-1} }$

$\longmapsto\rm{u = ( 1 \times 5 ) \; ms^{-1} }$

$\longmapsto\bf{u = 5 \; ms^{-1} }$

Converting final velocity in m/s :

$\longmapsto\rm{v = 72 \; kmh^{-1} }$

$\longmapsto\rm{v =\Bigg ( 72 \times \dfrac{5}{18} \Bigg ) \; ms^{-1} }$

$\longmapsto\rm{v= ( 4 \times 5 ) \; ms^{-1} }$

$\longmapsto\bf{v = 20 \; ms^{-1} }$

Now, we have to find the acceleration, so by using the acceleration formula, we have :

$\longmapsto\bf{a = \dfrac{v-u}{t} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\longmapsto\rm{a = \dfrac{20 -5}{10}\; ms^{-2} }$

$\longmapsto\rm{a = \dfrac{15}{10}\; ms^{-2} }$

$\longmapsto\bf{a = 1.5 \; ms^{-2} }$

Now, we have acceleration, by using the third equation of motion, we have :

$\longmapsto\bf{v^2 - u^2 = 2as }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

$\longmapsto\rm {(20)^2 - (5)^2 = 2\times 1.5 \times s }$

$\longmapsto\rm {400 - 25 = 3s }$

$\longmapsto\rm {375= 3s }$

$\longmapsto\rm {\cancel{\dfrac{375}{3}}= s }$

$\longmapsto\bf {125 \; m= s }$

∴ Distance travelled is 125 m.