Question

A car travels with a uniform velocity of
30ms for 6 s. The brakes are then applied
and the
car is uniformly retarded. It
comes to rest in further los.
a.Draw a graph of velocity and time.
b) Find total distance travelled by the car
at the end of journey
​

Answer 1

GiveN :

• Initial velocity (u) = 30 m/s
• Time interval (t) = 6s
• Final velocity (v) = 0 m/s

To FinD :

• Distance travelled by car after applying breaks.

SolutioN :

Use 1st equation of motion :

$\implies \rm{v = u + at}$

ㅤㅤㅤㅤㅤㅤㅤ

$\implies \rm{0 = 30 + 6a}$

ㅤㅤㅤㅤㅤㅤㅤ

$\implies \rm{6a = -30}$

ㅤㅤㅤㅤㅤㅤㅤ

$\implies \rm{a = \dfrac{-30}{6}}$

ㅤㅤㅤㅤㅤㅤㅤ

$\implies \rm{a = -5}$

ㅤㅤㅤㅤㅤㅤㅤ

$\underline{\rm{\therefore \: Retardation \: is \: -5 \: ms^{-2}}}$

____________________________

Now, use 3rd equation of motion :

$\implies \rm{v^2 - u^2 = 2as}$

ㅤㅤㅤㅤㅤㅤㅤ

$\implies \rm{0^2 - 30^2 = 2(-5)s}$

ㅤㅤㅤㅤㅤㅤㅤ

$\implies \rm{0 - 3600 = -10s}$

ㅤㅤㅤㅤㅤㅤㅤ

$\implies \rm{10s = 3600}$

ㅤㅤㅤㅤㅤㅤㅤ

$\implies \rm{s = \dfrac{3600}{10}}$

ㅤㅤㅤㅤㅤㅤㅤ

$\implies \rm{s = 360}$

ㅤㅤㅤㅤㅤㅤㅤ

$\small \underline{\rm{\therefore \: Distance \: travelled \: by \: car \: before \: coming \: to \: rest \: is \: 360 \: m}}$

Answer 2

• Given -

• A car travels with a uniform velocity of 30m/s for 6sec. Then brakes are applied & car is uniformly retarted
• Initial Velocity = 30m/s
• Time = 6sec
• Final Velocity = 0m/s ... (Brakes are applied. No motion)

• To Find -

• Distance covered by car after applying brakes

• Solution -

First of all, Apply 1st eqn of motion to find out Retardation.

Placing values according to given Conditions,

$⟹v = u + at$

$⟹0 = 30 + 6a$

$⟹6a = 0 - 30$

$⟹6a = - 30$

$⟹a = \frac{\cancel{-30}}{\cancel{6}}$

$⟹a = - 5$

⛬ Retardation of car is -5 m/s^2

Applying these value in 3rd eqn of motion to find out Distance,

$⟹ {v}^{2} - {u}^{2} = 2as$

$⟹ {0}^{2} - {30}^{2} = 2( - 5)s$

$⟹0 - 3600 = - 10s$

$⟹10s = 3600$

$⟹s = \frac{\cancel{3600}}{\cancel{10}}$

$⟹s = 360$

⛬ Distance covered by car after applying brakes is 360m