Question

A bus starting from rest moves with a uniform acceleration of 0.2 m s-2 for 5 minutes. calculate the speed at wired and the distance covered​

Answer 1

Answer: YOUR ANSWER

Explanation: Initial velocity, u=0m/s  

Final velocity, yv=?

Acceleration, a=0.2m/s  

2

 

Time, t=5min=5×60=300sec

Using first equation of motion to obtain the final speed:

v=u+at

v=0+0.2×300=60m/s

And the distance travelled is  

s=ut+  

2

1

​

αt  

2

 

s=0×300+  

2

1

​

×0.2×300×300

s=0+9000=9000m=9km

Answer 2

Answer:

Given :-

• A bus started from rest moves with a uniform acceleration of 0.2 m/s² for 5 minutes.

To Find :-

• What is the speed acquired and the distance covered.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{v =\: u + at}}}\: \: \: \bigstar$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

$\clubsuit$ Third Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{v^2 - u^2 =\: 2as}}}\: \: \: \bigstar$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance Covered

Solution :-

First, we have to find the final velocity or the speed acquired :

Given :

• Acceleration = 0.2 m/s²
• Time taken = 5 minutes = 300 seconds
• Initial Velocity = 0 m/s

According to the question by using the formula we get,

$\implies \bf v =\: u + at$

$\implies \sf v =\: 0 + (0.2)(300)$

$\implies \sf v =\: 0 + 0.2 \times 300$

$\implies \sf v =\: 0 + 60$

$\implies \sf\bold{\red{v =\: 60\: m/s}}$

Hence, the speed or final velocity is 60 m/s .

Now, we have to find the distance covered :

Given :

• Initial Velocity = 0 m/s
• Final Velocity = 60 m/s
• Acceleration = 0.2 m/s²

According to the question by using the formula we get,

$\implies \bf v^2 - u^2 =\: 2as$

$\implies \sf (60)^2 - (0)^2 =\: 2(0.2)s$

$\implies \sf 3600 - 0 =\: 0.4s$

$\implies \sf 3600 =\: 0.4s$

$\implies \sf \dfrac{3600}{0.4} =\: s$

$\implies \sf 9000 =\: s$

$\implies \sf\bold{\red{s =\: 9000\: m}}$

$\therefore$ The speed acquired is 60 m/s and the distance covered by the bus is 9000 m .