Question

1) A car accelerates uniformly from 18km/hr to 36 km/hr in 5s.
a) The acceleration and,
b) the distance covered by the car in that time.

2) For a motion, u = 12m/sec, a = 2m/sec², T = 15 sec, Find V ?

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Answer 1

1)

GIVEN

• Initial Velocity ( u ) = 18 km/h = 18 / 3.6 = 5 m/s

• Final Velocity ( v ) = 36 km/h = 36 / 3.6 = 10 m/s

• Time = 5 second

TO FIND

• The acceleration

• The distance travelled by the car in that time

We know that,

• Acceleration ( a ) = ( v - u ) / t

• Acceleration ( a ) = ( 10 - 5 ) / 5

• Acceleration ( a ) = 5 / 5

• Acceleration ( a ) = 1

• Acceleration ( a ) = 1 m/s²

Acceleration ( a ) = 1 m/s²

We know that,

• Distance = u × t + ½ × a × t²

• Distance = 5 × 5 + ½ × 1 × 5²

• Distance = 25 + 0.5 × 25

• Distance = 25 + 12.5

• Distance = 37.5

• Distance = 37.5 m

Distance = 37.5 m

___________________________

2)

GIVEN

• Initial Velocity ( u ) = 12 m/s

• Acceleration ( a ) = 2 m/s²

• Time ( t ) = 15 sec

TO FIND

• Final Velocity ( v )

We know that,

• Final Velocity ( v ) = u + a × t

• Final Velocity ( v ) = 12 + 2 × 15

• Final Velocity ( v ) = 12 + 30

• Final Velocity ( v ) = 42

• Final Velocity ( v ) = 42 m/s

Final Velocity ( v ) = 42 m/s

Answer 2

${ \underline{ \pink{ \rm{Solution (1)}}}}$

${\underline{\boxed{\mathcal{\pmb{\quad GIVEN \qquad}}}}}$

• u (initial velocity) = 18km/hr
• v (final velocity) = 36 km/hr
• t (Time) = 5s.

${\underline{\boxed{\mathcal{\pmb{\quad FIND\: \: OUT \qquad}}}}}$

• a (Acceleration) ?
• The distance covered by the car in that time?

${\underline{\boxed{\mathcal{\pmb{\quad SOLUTION \qquad}}}}}$

$\implies{ \rm{ \cancel{18 } \times \frac{5}{ \cancel{18}} = 5 \: m/s}}$

$\implies{ \rm{ {36} \times \frac{5}{ 18} = 10\: m/s}}$

Now, We have :-

• u (initial velocity) = 5m/s

• v (final velocity) = 10 m/s

• t (Time) = 5s.

In first part :-

We know that,

$\star{ \underline{ \boxed{ \green{ \rm{Acceleration = \frac{v - u}{t} }}}}}$

• Now, substitute the values :-

$\implies{ \rm{a = \frac{10m/s^{ - 1 } - 5m/s ^{ - 1} }{5 \: s} }}$

$\implies{ \rm{a = \frac{5m/s ^{ - 1} }{5 \: s}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\implies{ \boxed{ \rm{a = 1 \: m/s ^{ - 2}}} }$

In Second part :-

We know that,

${ \underline{ \boxed{ \red{ \rm{s = ut + \frac{1}{2} at ^{2} }}}}}$

• Now, Substitute the values :-

$\implies{ \rm{s = 5 \times 5 + \frac{1}{2} \times 5^{2} }}$

$\implies{ \rm{s = 25 + \frac{1}{2} \times 5 ^{2}}}$

$\implies{ \rm{s = 25 + 12.5}}$

$\implies{ \boxed{ \rm{s = 37.5 \: m}}}$

• Hence, the acceleration of the car is ${ \rm{1 \: m /s ^{ - 2} }}$ and the distance covered is 37.5 m

${ \underline{ \pink{ \rm{Solution (2)}}}}$

${\underline{\boxed{\mathcal{\pmb{\quad GIVEN \qquad}}}}}$

• u (initial velocity) = 12 m/s
• a (Acceleration) = 2 m/s²
• t (Time) = 15s

${\underline{\boxed{\mathcal{\pmb{\quad FIND\: \: OUT \qquad}}}}}$

• v (Final velocity) ?

${\underline{\boxed{\mathcal{\pmb{\quad SOLUTION \qquad}}}}}$

We know that,

${ \underline{ \boxed{ \purple{ \rm{v = u + at}}}}}$

• Now, Substitute the values :-

$\implies{ \rm{v = 12 + 2 \times 15}}$

$\implies{ \rm{v = 12 + 30}}$

$\implies{ \boxed{ \rm{ v = 42m/s}}}$

• Hence, v (Final velocity) is 42 m/s.

____________________________

${ \underline{\blue{ \rm { More \: Information :}}}}$

★ What is Speed ?

• The travelled by the object in unit time.
• The SI unit of speed is m/s.

★ What is Velocity ?

• The rate of motion of an object can be more comprehensive if we specify its direction of motion along with its speed the quantity that specifies both these aspects is called velocity.

★ What is Acceleration ?

• Acceleration is the rate of change of velocity.
• The SI unit of acceleration is m/s².

★ Equations of motion by Graphical method.

• When an object moves along a straight line with uniform acceleration, its possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations are given below :

1. ${ \rm{v = u + at}}$
2. ${\rm{s = ut + \frac{1}{2} at ^{2}}}$
3. ${\rm{2as = v ^{2} - u^{2}}}$

★ Where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the Final velocity, and s, is the distance travelled by the object in time t.