Question

Q1.A motorcycle moving with speed of 18km/h is subjected to an acceleration of 0.2 metre per second square. Calculate the speed of the motorcycle after 10 seconds, and the distance travelled in this time.

Answer 1

$\huge{\underline{\mathfrak{❥Given/That:-}}}$

Given that

The initial speed of car=u= 5 m/s

Acceleration =a=0.2m/s2

Time=t=10s

We need to calculate the distance

We know that

v = u + at

= 5 + 0.2×10

= 5+2

= 7 m/s

s = ut + 1/2at²

= (5×10) + (1/2)×(2/10)×(10×10)

= 50 + 10

= 60 m

Final speed = 7 m/s

2) Given that

Initial velocity = 18 km/h

We will express it in m/s

So Initial velocity = 18 km/h= 18×5/18 m/sec= 5 m/sec

The final velocity = zero

Acceleration = change in velocity/time

= ( final velocity – intial velocity)/time

= ( 0 – 5)/2.5 = -2 m/s²

so, retardation = 2m/s²

Explanation:

$\huge{\underline{\mathfrak{❥Sachuuu❣}}}$

Answer 2

Explanation:

✬ Final Velocity = 7 m/s ✬

✬ Distance = 60 m ✬

Explanation:

Given:

• Initial velocity of motorcycle is 18 km/h.
• Acceleration is 0.2 m/s².
• Time is 10 seconds.

To Find:

• Final velocity of motorcycle and Distance travelled

Formula to be used:

v = u + at (for final velocity)

s = ut + 1/2at² (for distance)

Solution: Changing the initial velocity into m/s.

➟ u = 18(5/18)

➟ u = 5 m/s

Now put the values on formula

⟹ v = u + at

⟹ v = 5 + 0.2(10)

⟹ v = 5 + 2

⟹ v = 7

So, Final velocity of bike is 7 m/s.

⟹ s = ut + 1/2at²

⟹ s = 5(10) + 1/2(0.2)(10)²

⟹ s = 50 + 0.2/2(100)

⟹ s = 50 + 20/2

⟹ s = 50 + 10

⟹ s = 60

Hence, the distance travelled is 60 m.