Question

A bus starting from rest moves with a uniform acceleration of 0.1 ms$^{–2}$ for 2 minutes. find speed acquired and the distance travelled​

Answer 1

Explanation:

HEY MATE .........

$a = 0.1m | {s}^{2} | \\ \\ and \: t = 2min \: = 120 |sec| \\ \\ a.t.q \\ \\ v = u + at \\ \\ = 0 + 0.1 \times 120 \\ \\ v = 12m |s| \\ \\ \\ b)s = ut + \frac{1}{2} a {t}^{2} \\ \\ s = 0 + \frac{1}{2} \times 0.1 \times ( {120}^{2} ) \\ \\ s = 720 |m|$

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Answer 2

Answer:

Speed acquired is 12 m/s and the distance travelled is 720 m.

Explanation:

Given:

• Acceleration (a) = 0.1 m/s$^{2}$ .
• Time (t) = 2 min = 2 × 60 = 120 s.

To find:

Speed acquired and the distance travelled.

☯️ Formula used:

• First equation of motion:

v = u + at

where,

• v = final velocity
• u = initial velocity
• a = acceleration and
• t = time

• s = ut + 1/2at$^{2}$

where,

• s = distance
• u = initial velocity
• t = time and
• a = acceleration.

Solution:

According to the first equation of motion

v = u + at

Now substitute the values, we get

➳ v = 0 + 0.1 × 120

➳ v = 0 + 12 = 12 m/s.

$\therefore$ speed acquired = 12 m/s.

Now moving to next question,

we need to find the distance travelled.

s = ut + 1/2at$^{2}$

➳ s = 0 + 1/2 × 0.1 × (120s)$^{2}$

➳ s = 0 + 1/2 × 1440 m

➳ s = 1/2 × 1440 m

➳ s = 720 m.

Thus, the distance travelled is 720 m.