Question

Q] A motorcycle moving with a speed of 5 m/s is subjected to an acceleration of 0.2
ms- 2 . Calculate the speed of the motor cycle after 10 second, and the distance
travelled in this time

Answer 1

━━━━━━━━━━━━━━━━━━━━

✺Answer:

♦️GiveN

• Started with velocity 5 m/s
• Accelerated at the rate 0.2 m/s^2
• Time taken is 10 s.

♦️To FinD

• Speed of body after 10 s
• Distance travelled in this time interval.

━━━━━━━━━━━━━━━━━━━━

✺We must know:

Before solving the question, we must be known to certain important things like equations of motion.

Whenever a body is acclerated at a uniform rate, we can use the Equations of motion but if is non uniform, then we need to do it with calculus.

Here, Body is showing uniform Accleartion, so we can apply the three equations of motion which are:-

$\large{ \begin{cases} \: \rm{v = u + at...............(1) } \\ \rm{s = ut + \frac{1}{2}a {t}^{2}..............(2) } \\ \rm{{v}^{2} - {u}^{2} = 2as}..............(3) \end{cases}}$

♦️Note:

Symbols have their usual meaning.

━━━━━━━━━━━━━━━━━━━━

✺Solution:

We have,

Initial velocity (u)= 5 m/s

Accleration (a) = 0.2 m/s^2

Time taken (t) = 10 s

By using 1st equation of motion,

$\large{ \rm{ \rightarrow v = 5 + 0.2 \times 10 }} \\ \\\large{ \rm{ \rightarrow v = 5 + 2}} \\ \\ \large{ \boxed{ \red{ \rm{ \rightarrow v = 7 \: m {s}^{ - 1}}}}}$

So, Final velocity = 7 m/s

━━━━━━━━━━━━━━━━━━━━

Now, For finding distance, we can use 2nd eqn. of motion,

By using 2nd equation of motion,

$\large{ \rm{ \rightarrow s = 5 \times 10 + \frac{1}{ \cancel{2} } \times \cancel{0.2 } \: \: 0.1 \: \times {10}^{2} }} \\ \\ \large{ \rm{ \rightarrow s = 50 + \frac{100}{10} \: \: m}} \\ \\\large{ \boxed{ \red{ \rm{ \rightarrow s = 60 \: m}}}}$

▶️Or.....

As we have calculated final velocity, we can use the 3rd eq.n of motion also.

So By using 3rd equation of motion,

$\large{ \rm{ \rightarrow {7}^{2} - {5}^{2} = 2 \times 0.2 \times s}} \\ \\ \large{ \rm{ \rightarrow 49 - 25 = 0.4 \times s}} \\ \\\large{ \rm{ \rightarrow s = \frac{24}{0.4} }} \\ \\ \large{ \rm{ \rightarrow s = \cancel{ \frac{24 \times 10}{4} }}} \\ \\ \large{ \boxed{ \red{ \rm{ \rightarrow s = 60 \: m}}}}$

So, Distance covered By Body = 60m

━━━━━━━━━━━━━━━━━━━━

✺So Final Answer:

$\large{ \rm{ \therefore{ \purple{ \: final \: velocity \: of \: body = 7 \: m {s}^{ - 1} }}}} \\ \\ \large{ \rm{ \therefore{ \purple{distance \: covered \: by \: the \: body = 60 \: m}}}}$

━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

Final velocity = 7 m/s

Final velocity = 7 m/sDistance = 60 metres

Explanation:

Given :

Initial velocity (u) = 5 m/s

Acceleration (a) = 0.2 m/s^2

Time (t) = 10 seconds

To find :

• Final velocity (v)
• Distance (s)

We can find the final velocity with the first equation of motion which says :

V=u+at

V = 5+0.2×10

V=5+2

V=7 m/s

The speed of the motor cycle after 10 seconds will be 7 m/s

We can find the distance with the third equation of motion which says :

V^2-u^2=2as

7^2 - 5^2 = 2×0.2×s

49-25 = 0.4s

24 = 0.4s

S=60m

The distance covered will be 60 metres