Question

of the taxi during the entire journey.
4. At a certain instant, an auto driver is driving at a speed of 30 km h-1. Then, he uniformly accelerates the auto
and after 2.5 minutes, the speed of his auto becomes 60 km h-1. What is the average speed of the auto? What is
the value of uniform acceleration during the given interval of time?
Laser signal ent from the earth takes a ne of 1.28 seconds to reach the earth. If the speed of laser signal be​

Answer 1

Answer:

Acceleration = 1/18 m/s^2

Average speed = 12.5 m/s

Explanation:

Given :

Initial velocity = u = 30 km/h

30 km/h = 30×5/18 = 25/3 m/s

Time = t = 2.5 minutes

1 minute = 60 seconds

2.5 minutes = 2.5×60 = 150 seconds

Final velocity = v = 60 km/h

60 km/h = 60×5/18 = 50/3 m/s

To find

• Acceleration
• Average speed

We can find acceleration with first equation of motion which says :

V = u+at

50/3=25/3+150a

25/3 = 150a

25/3 × 1/150 = a

a = 1/18 m/s^2

a = 0.05555... m/s^2

The acceleration is equal to 0.055555... m/s^2

Average speed = total distance / total time

S=ut+1/2at^2

S=25/3×150+1/2×1/18×22500

S=1250+625

S=1875 metres

Average speed = 1875/150 = 12.5 m/s

Answer 2

Given:-

Initial velocity of auto(u) =30km/h

Final velocity (v)= 60km/h

Time (t) = 2.5 minutes

To find :-

• Acceleration

• Average speed

• First find the acceleration

$\boxed{\sf{ v= u+at}}$

v= 60km/h in m/s

$\sf \ \ 60\times \dfrac{5}{18} = \cancel{\dfrac{300}{18}} = \dfrac{50}{3}m/s$

u= 30km/h

$\sf \ \ 30\times \dfrac{5}{18} = \cancel{\dfrac{150}{18}} = \dfrac{25}{3}m/s$

T= 2.5 minutes

=> 2.5×60= 150 seconds

Acceleration ?

$\longrightarrow\sf v=u+at\\ \\ \longrightarrow\sf \dfrac{50}{3}= \dfrac{25}{3}+a\times 150\\ \\ \longrightarrow\sf \dfrac{50}{3}-\dfrac{25}{3}= 150\times a\\ \\ \longrightarrow\sf \dfrac{25}{3}= 150\times a\\ \\ \longrightarrow\sf \dfrac{\dfrac{25}{3}}{150}=a\\ \\ \longrightarrow\sf \dfrac{\cancel{25}}{3}\times \dfrac{1}{\cancel{150}}=a\\ \\ \longrightarrow\sf \dfrac{1}{3\times 6}= a\\ \\ \longrightarrow\sf a= \dfrac{1}{18}= 0.556m/s^2$

$\boxed{\bigstar{\sf{ Acceleration = \dfrac{1}{18}\ \ or \ \ 0.055m/s^2}}}$

Now we have to find the average speed

• First find the Distance !
• By using third equation of motion

$\boxed{\sf{\dag\ \ v^2=u^2+2as}}$

$\longrightarrow\sf \bigg(\dfrac{50}{3}\bigg)^2= \bigg(\dfrac{25}{3}\bigg)^2 +\cancel{2}\times \dfrac{1}{\cancel{18}}\times s\\ \\ \longrightarrow\sf \dfrac{2500}{9} -\dfrac{625}{9}= \dfrac{s}{9} \\ \\\longrightarrow\sf \dfrac{1875}{\cancel{9}}\times \cancel{9}= s\\ \\ \longrightarrow\sf 1875 = s$

$\boxed{\sf{\dag\ \ Distance = 1875m}}$

• Find the average speed

$\longrightarrow\sf Average\ speed = \dfrac{Total \ Distance }{Total \ Time}\\ \\ \longrightarrow\sf Av.\ speed = \dfrac{1875}{150}\\ \\ \longrightarrow\sf Av.\ speed = 12.5m/s$

$\boxed{\bigstar{\sf{Average \ speed = 12.5m/s}}}$