Question

A bus starting from rest acquires a uniform acceleration of 0.1m/s^2 for 2 min.Find the a) speed and b)distance travelled.

Answer 1

Answer:

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a = 0.1 m/s²

t = 120 sec

u = 0 m/sec

v = u + at

v = 12 m/sec

s = ut + 1/2at²

= 1/2(0.1) × 120 × 120

= 720 meters

Answer 2

Given

A bus starting from rest acquires a uniform acceleration of 0.1m/s² for 2 min

To Find

a. Speed

b. Distance travelled

Key Points

SI unit of speed is m/s

SI unit of time is second

SI unit of distance is meter

SI unit of acceleration is m/s²

Equations of motion :

$\bigstar\ \; \bf \pink{v=u+at}\\\\\bigstar\ \; \bf \green{s=ut+\dfrac{1}{2}at^2}\\\\\bigstar\ \; \bf \blue {v^2-u^2=2as}\\\\\bigstar\ \; \bf \red {S_n=u+\dfrac{a}{2}(2n-1)}$

where ,

• u denotes initial velocity
• v denotes final velocity
• a denotes acceleration
• t denotes time
• s denotes distance traveled
• Sₙ denotes distance travelled in nth second
• n denotes nth second

Solution

Given ,

• u = 0 m/s
• a = 0.1 m/s²
• t = 2 min = 2 × 60 = 120 s

Apply 1st equation of motion ,

⇒ v = u + at

⇒ v = 0 + (0.1)(120)

⇒ v = 0 + 12

⇒ v = 12 m/s

So , Speed = 12 m/s  (a)

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (12)² - (0)² = 2(0.1)s

⇒ 144 = 2(0.1)s

⇒ 0.1s = 72

⇒ s = 720 m

So , Distance traveled = 720 m (b)