Question

as bus starting from rest moves with uniform acceleration 0.1 metre per second square for 2 minutes. then it moves with a uniform speed for next 5 minutes. then the driver applies brakes and bus comes to rest in another 1 minute.. calculate the total distance covered by bus and average speed​

Answer 1

Answer:

Total distance covered = 4680 m

Average speed = 9.75 m/s

Explanation:

Given :

• A bus starting from rest moves with uniform acceleration 0.1 m/s² for 2 minutes.

• Then it moves with a uniform speed for next 5 minutes.
• Then the driver applies brakes and bus comes to rest in another 1 minute.

To find :

the total distance covered by bus and average speed​

Solution :

First, let's find the distance covered by bus in first 2 minutes.

It started from rest, hence initial velocity, u = 0 m/s

acceleration, a = 0.1 m/s²

time, t = 2 min = 2 × 60 sec = 120 sec

From the second equation of motion,

$\bf S=ut+\dfrac{1}{2}at^2$

S = (0)(120) + [ (1/2) × 0.1 × 120² ]

S = 0 + [ 1/20 × 14400 ]

S = 720 m

Next, it moves with uniform speed for next 5 minutes.

As moving with uniform speed, acceleration = 0 m/s²

• speed = distance/time

time = 5 min = 5 × 60 sec = 300 sec

we have to find the speed.

required speed = speed at the end of first 2 min,

From the first equation of motion,

v = u + at

v = 0 + (0.1) (120)

v = 12 m/s

∴ The required speed = 12 m/s

Speed = distance/time

distance = speed × time

distance = 12 m/s × 300 sec

distance = 3600 m

The driver applies brakes and bus comes to rest in another 1 minute.

In this case,

initial velocity, u = 12 m/s

final velocity, v = 0 m/s (since it comes to rest)

time, t = 1 min = 60 sec

From the first equation of motion,

v = u + at

0 = 12 + a(60)

60a = -12

a = -12/60

a = -0.2 m/s²

Substitute the values in v² - u² = 2as

0² - 12² = 2(-0.2)s

-144 = -0.4 × s

 s = 144/0.4

 s = 1440/4

 s = 360 m

∴ The total distance covered by bus = 720 m + 3600 m + 360 m

       = 4680 m

total time taken = 2 min + 5 min + 1 min = 8 min = 480 sec

$\sf Average \ speed=\dfrac{total \ distance}{total \ time \ taken}$

Average speed = 4680 m/480 s

Average speed = 9.75 m/s

Answer 2

Explanation:

Total distance covered = 4680 m

Average speed = 9.75 m/s

Explanation:

Given :

A bus starting from rest moves with uniform acceleration 0.1 m/s² for 2 minutes.

Then it moves with a uniform speed for next 5 minutes.

Then the driver applies brakes and bus comes to rest in another 1 minute.

To find :

the total distance covered by bus and average speed

Solution :

First, let's find the distance covered by bus in first 2 minutes.

It started from rest, hence initial velocity, u = 0 m/s

acceleration, a = 0.1 m/s²

time, t = 2 min = 2 × 60 sec = 120 sec

From the second equation of motion,

\bf S=ut+\dfrac{1}{2}at^2S=ut+

2

1

at

2

S = (0)(120) + [ (1/2) × 0.1 × 120² ]

S = 0 + [ 1/20 × 14400 ]

S = 720 m

Next, it moves with uniform speed for next 5 minutes.

As moving with uniform speed, acceleration = 0 m/s²

speed = distance/time

time = 5 min = 5 × 60 sec = 300 sec

we have to find the speed.

required speed = speed at the end of first 2 min,

From the first equation of motion,

v = u + at

v = 0 + (0.1) (120)

v = 12 m/s

∴ The required speed = 12 m/s

Speed = distance/time

distance = speed × time

distance = 12 m/s × 300 sec

distance = 3600 m

The driver applies brakes and bus comes to rest in another 1 minute.

In this case,

initial velocity, u = 12 m/s

final velocity, v = 0 m/s (since it comes to rest)

time, t = 1 min = 60 sec

From the first equation of motion,

v = u + at

0 = 12 + a(60)

60a = -12

a = -12/60

a = -0.2 m/s²

Substitute the values in v² - u² = 2as

0² - 12² = 2(-0.2)s

-144 = -0.4 × s

s = 144/0.4

s = 1440/4

s = 360 m

∴ The total distance covered by bus = 720 m + 3600 m + 360 m

= 4680 m

total time taken = 2 min + 5 min + 1 min = 8 min = 480 sec

\sf Average \ speed=\dfrac{total \ distance}{total \ time \ taken}Average speed=

total time taken

total distance

Average speed = 4680 m/480 s

Average speed = 9.75 m/s