Question

time 4 sec, velocity 36 km/ hr,mass 700 kg : questions) ,find the distance covered with in time ,the force applied to Stop the bus,) answer is 20 m,- 17500 N step wise answer ❤️❤️​

Answer 1

Answer:

• Distance Travelled = 20 metres
• Force required to stop the bus = 1750 Newton

Explanation:

Given:-

• Time ,t = 4 s
• Initial velocity ,u = 36km/h = 36×5/18 = 10m/s
• Final velocity ,v = 0m/s
• Mass, m = 700 kg

To Find:-

• Forced applied to stop the bus ,F
• Distance travelled ,s

Solution:-

Firstly we calculate the acceleration of the bus . Using 1st Equation of Motion.

• v = u+at

where,

• v is the final velocity
• a is the acceleration
• u is the initial velocity
• t is the time taken

Substitute the value we get

→ 0 = 10 + a×4

→ -10 = a×4

→ a = -10/4

→ a = -2.5 m/s²

Here negative sign show retardation . acceleration of the bus is 2.5m/s².

Now, calculating the distance travelled by the bus. Using 3rd Equation of Motion

• v² = u² +2as

Substitute the value we get

→ 0² = 10² + 2×(-2.5) ×s

→ 0 = 100 + (-5) ×s

→ -100 = -5×s

→ s = -100/-5

→ s = 20m

• Hence, the distance travelled by the bus is 20 metres.

Now, we have to calculate the force applied to stop the bus .

• F = ma

where,

• F is the force
• m is the mass
• a is the acceleration

Substitute the value we get

→ F = 700×2.5

→ F = 70×25

→ F = 1750 N

• Hence, the force required to stop the bus is 1750 Newton.

Answer 2

Answer:

Given :-

• Time taken = 4 sec
• Final velocity = 36 km/hr × 5/18 = 10 m/s
• Mass = 700 kg

To Find :-

• Force applied

Solution :-

Firstly let's find Acceleration by using first equation

As we know that

$\huge \fbox{V = u + at}$

Here,

V = Final Velocity = 10 m/s

U = Initial velocity = 0 m/s

A = Acceleration = ?

T = Time = 4 sec

$\tt \: 10 = 0 + a \times 4$

$\tt \: 0 - 10 = 4a$

$\tt \: - 10 = 4a$

$\tt \: a = \dfrac{ - 10}{4} = 2.5$

Hence :-

Acceleration of body is 2.5 m/s²

Now,

Let's find Distance travelled by using third Equation

$\huge \bf \: {v}^{2} - {u}^{2} = 2as$

Here,

V = Final Velocity

U = Initial velocity

A = Acceleration

S = Distance

$\tt \: {10}^{2} - {0}^{2} = 2( - 2.5)(s)$

$\tt \: 0 - 100 = - 5s$

$\tt \: - 100 = - 5s$

$\tt \: s \: = \dfrac{ - 100}{ - 5} = 20 \: m$

Hence :-

Distance travelled by body is 20 m

Now,

Let's find Force applied

As we know that

$\huge \fbox{F = ma}$

Here,

F = Force Applied

M = Mass

A = Acceleration

$\tt \:F = 700 \times 2.5$

$\tt \: F = 1750$

Hence :-

Force applied is 1750 N.