Question

a bus starting from rest moves with a uniform acceleration of 0.1 metre per second square 45 minutes find velocity acquired and distance travelled​

Answer 1

Given :–

• Acceleration (a) = 0.1 m/s²
• Time (t) = 45 mins.

To Find :–

• Velocity acquired.
• Distance travelled.

Solution :–

Given that,

• a = 0.1 m/s²
• t = 45 mins.

We need to convert minutes into seconds.

For converting we have to multiply the time value by 60.

So,

→ 45 × 60s

→ 2700s

Now,

1. Velocity acquired (v).

We know that,

$\boxed{v = u + at}$

Here,

• u = 0.
• a = 0.1 m/s².
• t = 2700s.

$\mapsto v = 0 + (0.1 \: \frac{m}{ { \cancel{s}}^{2} } )(2700 \: \cancel{s})$

$\mapsto v = 0 + (0.1 \: \dfrac{m}{s} )(2700 \: s)$

$\mapsto v = 0 + 270 \: \dfrac{m}{s}$

$\mapsto v = 270 \: \dfrac{m}{s}$

2. Distance travelled (s).

We know that,

$\boxed{s = ut + \dfrac{1}{2} a{t}^{2} }$

Here,

• u = 0.
• t = 2700s.
• a = 0.1 m/s².

$\mapsto s = (0)(2700 \: s) + \dfrac{1}{ 2} (0.1 \frac{m}{ \cancel{{s}^{2}} } )( {2700}^{2} \: \cancel{{s}^{2} })$

$\mapsto s = 0 + \dfrac{1}{ \cancel2} \times 0.1 \: m \times \cancel{7290000}$

$\mapsto s = 0 + 0.1 \times 3645000 \: m$

$\mapsto s = 0 + 364500 \: m$

$\mapsto s = 364500 \: m$

Hence,

The velocity acquired is 270 m/s and the distance travelled is 364500 m.