Question

if a bus moves with the speed of72km/hr,after 3 seconds it travels 120m distance. calculate final belocity and acceleration of bus.​

Answer 1

✪ AnsWer : ✪

Given;

• Initial velocity (u) = 72 km/h = 72 × (5/18) = 360/18 = 20 m/s
• time interval (t) = 3 s
• distance (s) = 120 m

Unknown Quantities;

• Final velocity (v)
• Acceleration (a)

Solution;

By applying second kinematical equation of motion we have:

⇥s = ut + ½ at²

⇥120 = 20 × 3 + ½ (a)(3)²

⇥120 = 60 + 9/2 × a

⇥120 - 60 = 9/2 × a

⇥60 = 9/2 × a

⇥60 × 2 = 9a

⇥120/9 = a

⇥a = 13.4 m/s²

Now,by applying first kinematical equation of motion we have:

⇥v = u + at

⇥v = 20 + 13.4 × 3

⇥v = 20 + 40.2

⇥v = 60.2 m/s

Answer 2

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◉GIVEN:-

• Initial speed (u) :- 72km/hr

In m/s,

$\sf{72 \times \frac{5}{18} =20m {s}^{ - 1} }$

• Distance(d):- 120m

• Time interval (t) :- 3 seconds.

◉TO FIND:-

• final velocity (v)

• Acceleration (a)

◉SOLUTION:-

Using 2nd kinematics equation

$\sf{d = ut + \frac{1}{2} a {t}^{2} } \\ \\ : \longrightarrow\sf{120 = 20 \times 3 + \frac{1}{2} \times a \times {3}^{2} } \\ \\ : \longrightarrow\sf{120 = 60 + \frac{9a}{2} } \\ \\ : \longrightarrow\sf{120 - 60 = \frac{9a}{2} } \\ \\ : \longrightarrow\sf{60 = \frac{9a}{2} } \\ \\ : \longrightarrow\sf{a = \frac{120}{9} } \\ \\ : \longrightarrow\sf \pink{a = 13.3m {s}^{ - 2} }$

Now, Using 1st kinematics equation

$\sf{v \: = u \: + at} \\ \\ : \longrightarrow\sf{v = 20 + 13.3 \times 3} \\ \\ : \longrightarrow\sf{v = 20 + 39.9} \\ \\ : \longrightarrow\sf \pink{v = 59.9m {s}^{ - 1} }$

So,

Acceleration is 13.3m/sec^2

final velocity is 59.9m/sec^-1