Question

11. An engine is travelling along a circular railway track of radius 1500 metres with a speed
of 60 km/h. Find the angle in degrees turned by the engine in 10 seconds.​

Answer 1

Answer:

Angle turned in 10s = $(\dfrac{20}{\pi})^{\circ}$

Step-by-step explanation:

$\blacksquare$ We know that,

1)

$Speed = Radius \times Angular \ speed$

$v = r \omega$

2)

$Angular \ displacement = Angular \ speed \times Time$

$\theta = \omega t$

$\blacksquare$ Given that,

Radius = r = 1500m

Speed = v = 60 km/hr = 60 * 5/18 = $\dfrac{50}{3}$ m/s

Time = t = 10s

Substituting the value of 'v' and 'r' in 1) we get,

$\dfrac{50}{3}$ = 1500 * ω

ω = $\dfrac{1}{90}$ rad/s

Sustituting the value of 'ω' and 't' in 2) we get,

θ = $\dfrac{1}{90}$ * 10

θ = $\dfrac{1}{9}$ rad

θ = $\dfrac{1}{9} * \dfrac{180}{ \pi}=(\dfrac{20}{\pi})^{\circ}$

Answer 2

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$\huge\red{\underline{{\bf Question : }}}$

✦ An engine is travelling along a circular railway track of radius 1500 metres with a speed of 60 km/h. Find the angle in degrees turned by the engine in 10 seconds.

$\huge\red{\underline{{\bf Answer : }}}$

✧ $\green{\underline{{\bf Given : }}}$

✦ An engine is travelling along a circular railway track of radius 1500 metres with a speed

of 60 km/h.

✧ $\green{\underline{{\bf To Find : }}}$

✦ Find the angle in degrees turned by the engine in 10 seconds.

✧ $\green{\underline{{\bf Solution : }}}$

$<font color= "blue">$

⟹ Radius of the track (r) = 1500 m

⟹ Speed of the train = 60 × 1000/3600 = 50/3 m/s

⟹ So, circumference of the path

= 2(22/7)(1500)

= 66000/7 m

⟹ So, distance travelled in 10 sec.

= 10 (50/3) m

= 500/3 m

⟹ So, angular displacement is

¤ = l/r

= (500/3) / 1500

= 1/9

$<font color= "black">$

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