Question

A car travels with uniform velocity of 25 m s-1 for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s. Find: (i) The distance which the car travels before the brakes are applied, (ii) Retardation and (iii) The distance travelled by the car after applying the brakes. ​

Answer 1

Given :

• A car travels with uniform velocity of 25 m s-1 for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s.

To find :

• The distance which the car travels before the brakes are applied
• Retardation
• The distance travelled by the car after applying the brakes.

Solution :

• Initial velocity (u) = 25m/s
• Time (t) = 5s

As we know that

→ Speed = distance/time

→ Distance = speed × time

→ Distance = 25 × 5

→ Distance = 125 m

Hence, distance travelled by car before applying brakes is 125 m

• Initial velocity (u) = 25m/s
• Time (t) = 10s
• Final velocity (v) = 0 (brakes applied)

According to first equation of motion

→ v = u + at

→ 0 = 25 + a × 5

→ 0 = 25 + 10a

→ 10a = - 25

→ a = -25/10 = - 2.5m/s

Minus shows retardation of car. Hence, retardation of car is 2.5m/s

• Initial velocity (u) = 25m/s
• Final velocity (v) = 0
• Acceleration (a) = - 2.5m/s

According to third equation of motion

→ v² = u² + 2as

→ (0)² = (25)² + 2 × (-2.5) × s

→ 0 = 625 - 5s

→ 5s = 625

→ s = 625/5

→ s = 125 m

Distance travelled by the car after applying the brakes is 125m

Answer 2

Question:

A car travels with uniform velocity of 25$\sf { m/s }$ for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s.

Find:

• (i) The distance which the car travels before the brakes are applied.

• (ii) Retardation

• (iii) The distance travelled by the car after applying the brakes.

Solutions:

☆The distance which the car travels before the brakes are applied.

${\underline {\underline {\boxed {\rm {\green { Given: }}}}}}$

• Speed = $\sf { 25m/s }$
• time ( t ) = $\sf { 5 s }$

${\underline {\underline {\boxed {\rm {\green { Formula: }}}}}}$

• ${\underline {\underline {\boxed {\tt {\red { Distance = speed \times time}}}}}}$

${\underline {\underline {\boxed {\rm {\green { Calculation: }}}}}}$

Before the brakes are applied , let s be the distance travelled.

According to formula, substituting values-

$\sf { : \implies s = 25 \times 5 }$

$\sf\purple { : \implies s = 125m }$

$\therefore$ The distance travelled by the car before the brakes applied was 125m.

_____________________________

☆Retardation

[ You can find retardation in this case by two methods, first one by acceleration formula and second one by using first equation of motion. So, here we'll do both methods for better understanding. ]

Method 1st ( By acceleration formula )

${\underline {\underline {\boxed {\rm {\green { Given: }}}}}}$

• initial velocity ( u ) = $\sf { 25m/s }$
• final velocity ( v ) = $\sf { 0 }$
• time ( t ) = 10s

${\underline {\underline {\boxed {\rm {\green { Formula: }}}}}}$

• ${\underline {\underline {\boxed {\tt {\red { a = \frac{v-u}{t} }}}}}}$

${\underline {\underline {\boxed {\rm {\green { Calculation: }}}}}}$

According to formula, substituting values-

$\sf { :\implies a = \dfrac{ 0 -25}{10} }$

$\sf { :\implies a = \dfrac{ -25}{10} }$

$\sf { :\implies a =\cancel { \dfrac{ -5}{2}} }$

$\sf\purple { :\implies a = -2.5m/{s}^{2} }$

$\therefore$ Retardation is 2.5 m/s²

_______...

Method 2nd ( By using first equation of motion )

${\underline {\underline {\boxed {\rm {\green { Given: }}}}}}$

• initial velocity ( u ) = $\sf { 25m/s }$
• final velocity ( v ) = $\sf { 0 }$
• time ( t ) = 10s

${\underline {\underline {\boxed {\rm {\green { Formula: }}}}}}$

• ${\underline {\underline {\boxed {\tt {\red{ v = u + at }}}}}}$

${\underline {\underline {\boxed {\rm {\green { Calculation: }}}}}}$

According to formula, substituting values-

$\sf { :\implies 0 = 25 + 10a }$

$\sf { :\implies a = \dfrac{-25}{10} }$

$\sf { :\implies a =\cancel { \dfrac{ -5}{2} } }$

$\sf\purple { :\implies a = -2.5m/{s}^{2} }$

$\therefore$ Retardation is 2.5 m/s²

___________________________

☆The distance travelled by the car after applying the brakes.

${\underline {\underline {\boxed {\rm {\green { Given }}}}}}$

• initial velocity (u ) =$\sf { 25 m/s}$
• final velocity ( v ) = 0 m/s
• acceleration ( a ) = $\sf {-2.5 m/{s}^{2} }$ [ as per we found it in question 2 ]

___..

Let s be the distance travelled after applying the brakes.

${\underline {\underline {\boxed {\rm {\green { Formula: }}}}}}$

• ${\underline {\underline {\boxed {\tt {\red { {v}^{2} - {u}^{2} = 2as }}}}}}$

${\underline {\underline {\boxed {\rm {\green { Calculation: }}}}}}$

According to formula, substituting values-

$\sf { :\implies {(0)}^{2} - {(25)}^{2} = 2 ( -2.5 )(s) }$

$\sf { :\implies 0 - 625 = -5(s) }$

$\sf { :\implies \cancel {- } 625 =\cancel { -} 5(s) }$

$\sf { :\implies s = \cancel {\dfrac{625}{5} }}$

$\sf\purple { :\implies s = 125m}$

$\therefore$ Distance Travelled by the car after applying breaks is 125 m.

___________________