Question

A car of mass 1000 kg travelling at 72 kmh-1 is uniformly brought to rest over a distance of 40 m. Find (a) the average acceleration ; (b) the average breaking force

Answer 1

Explanation:

Given:-

• Mass of the car = 1000kg

• Initial velocity (u) = 72km/h

• Distance covered (s) = 40m

To Find:-

• The average acceleration.

• The average breaking force.

Solution:-

Converting 72km/h to m/s

= $\rm 72 \times \dfrac{5}{18}$

= 20m/s

Initial velocity (u) = 20m/s

Final velocity (v) = 0m/s (The car comes to rest)

According to the 2nd equation of motion:-

$\rm \dashrightarrow {v}^{2} = {u}^{2} + 2as$

$\rm \dashrightarrow {0}^{2} = {(20)}^{2} + 2a(40)$

$\rm \dashrightarrow {0} = 400 + 80a$

$\rm \dashrightarrow 80a = - 400$

$\rm \dashrightarrow a= \dfrac{ - 400}{80}$

$\rm \dashrightarrow a= - 5$

$\underline{\boxed{\pink{\rm \therefore Average \: acceleration = -5m/s^{2}}}}$

Now, Average breaking force:-

Breaking force = Mass × Acceleration

= 1000 × -5

= -5000N

$\underline{\boxed{\purple{\rm \therefore Average\: breaking\: force = -5000N}}}$

Answer 2

Answer:

data

mass= 1000kg

velocity = 72 km/ h ( firstly we bring it in initial velocity si unit so 1 kg = 1000 gram and one hour 60* 36 = 3600 seconds so 72 * 1000 /3600 = 12m/ s)

final velocity = 0 m/ s

acceleration = ?

average breaking force = ?

distance = 40 m

solution :

according to Newton third law =

2AS = (vi)² _ vf²

2 A × 40 = 0²_ 20²

80 a= _ 400

a= _ 400/ 40

a= _ 5 m/s²

average breaking force = ma

= 1000 × _5

= _5000 n