Question

A bus of starting from rest moves with uniform aceelaration of 0.5 m/s2 for 2 minutes
Find (a) Speed
(b) Distance

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Answer 1

Answer:

Speed= 12m/s and Distance= 720 m

Step-by-step explanation:

By first equation of motion,v=u+at,we have

Speed(v) = initial velocity(u) + acceleration(a) × time(t)

Now, putting the values we get,

v = 0+0.5×2

v = 12 m/s

Now, By Second equation of motion, s=ut+1/2×at², we have

Distance(d)= initial velocity(u)× time(t) + 1/2×acceleration(a)×time(t)²

Now,putting the values, we get

s= 0×2+1/2×0.5×2²

s= 720m

Answer 2

Answer :

➥ Final velocity of a bus = 60 m/s

➥ Distance travelled by a bus = 1.05 km

Given :

➤ Intial velocity of bus (u) = 0 m/s

➤ Acceleration of a bus (a) = 0.5 m/s²

➤ Time taken by a bus (t) = 2 min

To Find :

➤ Final velocity of a bus (v) = ?

➤ Distance travelled by a bus (s) = ?

Solution :

Final velocity of a bus

From first equation of motion

$\tt{:\implies v = u + at}$

$\tt{:\implies v = 0 + 0.5 \times 120}$

$\tt{:\implies v = 0 + 60}$

$\bf{:\implies \underline{ \: \: \underline{ \purple{ \: \: v = 60 \: m/s \: \: }} \: \: }}$

Hence, the final velocity of a bus is 60 m/s.

Distance travelled by a bus

From second equation of motion

$\tt{:\implies s = ut + \dfrac{1}{2}a {t}^{2} }$

$\tt{:\implies s = 0 \times 60 + \dfrac{1}{ \cancel{2}} \times 0.5 \times \cancel{60} \times 60 }$

$\tt{:\implies s = 0 + 1 \times 0.5 \times 35 \times 60}$

$\tt{:\implies s = 0 + 0.5 \times 35 \times 60}$

$\tt{:\implies s = 0 + 17.5 \times 60}$

$\tt{:\implies s = 0 + 1050}$

$\bf{:\implies s = 1050 \: m}$

Convert into km :

$\tt{:\implies s = \cancel{ \dfrac{1050}{1000} }}$

$\bf{:\implies \underline{ \: \: \underline{ \purple{ \: \: s = 1.05 \: km \: \: }} \: \: }}$

Hence, the distance travelled by a bus is 1.05 km.

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Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as