Question

A car travels with a uniform velocity of 25 m per second for 5 seconds the brakes are then applied in the car is uniformly retarded and comes to rest in for the 10 seconds find the distance with the car travel before the brakes are applied the retardation and the distance travelled by the car after applying the brakes

Answer 1

Distance travelled by the car before the brakes are applied =
Velocity = Distance /Time
v= s/t
25 = s/5
s= 25/5
s=5 m
So, the distance is 5 m.
Distance travelled by car after brakes are applied =
Acceleration = final velocity - initial velocity /time taken
a= v-u/t
a= 0- 25/10
a= - 25/10
a= - 2.5 m/s ^2
So, retardation is - 2.5 m/s^2.

Answer 2

Answer:

$\huge \underline \mathfrak \green{hello \: mate}$

$\huge \underline \bold \red{solution}$

i) To find the distance of the car before

the brakes are applied,

$\bold {velocity = 25{m}^{ - 1}}$

$\bold{time = 5s}$

$\bold{distance = velocity \times time}$

$\: \: \: \: \: \: \: \: \: \: \bold { = 25 \times 5}$

$\: \: \: \: \: \: \: \: \: \: \bold \purple{ = 125m \: \: \: ..answer}$

ii) To find the retardation,

$\bold{final \: velocity \: = 0{m}^{ - 1}}$

$\bold{initial \: velocity \: = 25{m}^{ - 1}}$

$\bold{acceleration = \frac{final \: velocity - initial \: velocity}{time}}$

$\bold{acceleration = \frac{0 - 25}{10}}$

$\bold{acceleration = \frac{ - 25}{10}}$

$\bold \purple{acceleration = - 2.5{m}^{ - 2} \: \: ..ans}$

iii) To find the distance travelled by the

car after applying the brakes,

$\bold{initial \: velocity = 25{m}^{ - 1}}$

$\bold{time = 10s}$

$\bold{acceleration = - 2.5{m}^{ - 2}}$

$\bold \pink{by \: using \: the \: formula}$

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

$\bold{s \: = \: 25 \times 10 \: + \: \frac{1}{2} \times - 2.5 \times{10}^{2}}$

$\bold{s \: = \: 250 \: + \: \frac{ - 2.5 \times 100}{2}}$

$\bold{s \: = \: 250 \: + \: ( - 125)}$

$\bold{s \: = \: 250 - 125}$

$\bold \purple{s \: = \: 125m \: \: \: ....ans}$