Question

Acar has initial velocity of 72 km h It is accelerated at 2 ms Calculate the final velocity
and the distance covered after 3 seconds. What is the distance covered by the car in 3
second
Ans: 26 m s. 69 m​

Answer 1

Aɴsᴡᴇʀ

we have,

• initial velocity, u = 72 km/h

first we have to convert the unit of velocity into m/s

$\bold{\implies u = \frac{ 72 \times 1 \cancel{000}}{36 \cancel{0 0}} m {s}^{ - 1} } \\ \\ \bold{\implies u = \frac{ \cancel{72} \times 10}{ \cancel{36}} \:m {s}^{ - 1} \: \: \: \: } \\ \\ \bold{ \implies u = 2 \times 10 \: m {s}^{ - 1} \: \: \: \: \: \: \:} \\ \\ \bold{\implies u = 20 \: m {s}^{ - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$

_______________________________

so, now we have,

• initial velocity, u = 20m/s
• acceleration, a = 2 m/s²
• time taken, t = 3 sec
• final velocity, v = ?
• distance, s = ?

now, by first equation of motion :

$\boxed{ \red {\bold{at = v - u}}}$

$\bold{ : \implies 2 \times 3=v - 20 } \\ \\ \bold{ : \implies6 = v - 20 } \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \: v = 6 + 20} \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \: \boxed{ \pink {\bold{v = 26 \: m {s}^{ - 1} }}}}$

_______________________________

now, by second equation of motion:

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

$\bold{ : \implies s =20 \times 3 + \frac{1}{ \cancel{2}} \times \cancel{ 2} \times {3}^{2} } \\ \\ \bold{ : \implies s = 60 + 9} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \: \boxed{ \pink{\bold{ s = 69 \: m}}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

_______________________________

hence, final velocity of the car is 26 m/s and distance covered by the car is 69 m

_______________________________

☯︎ important :-

• first equation of motion : at = v - u
• second equation of motion : s = ut + ½ at²
• third equation of motion : v² = u² + 2as

Answer 2

Given :-

• Initial velocity = 72 km/h

                                = 72 × 5/18

                                = 20 m/s

• Acceleration = 2 m/s²

To Find :-

• Final velocity after 3 seconds.

• Distance covered after 3 seconds.

Solution :-

We can find velocity after 3 seconds by using first equation of kinematics.

$\underline{\boxed{\bf v = u+at}}$

$\longrightarrow \sf v = 20 + 3 \times 6 \\\\\longrightarrow v = 20+6 \\\\\longrightarrow \bf v = 26$

Velocity after 3 seconds :- 26 m/s

We can find distance covered after 3 seconds by using the second equation of kinematics .

$\underline{\boxed{\bf S = ut+ \dfrac{1}{2}at^2}}$

$\longrightarrow \sf S = 20 \times 3+ \dfrac{1}{2} \times 2 \times 3 ^2 \\\\\longrightarrow S = 60+3 \times 3 \\\\\longrightarrow S = 60+ 9 \\\\\longrightarrow \bf S = 69$

Distance covered after 3 seconds :- 69 m