Question

A motorcycle moving with a speed of 5m/s is subjected to an acceleration of 0.2 m/s∧2
Calculate the speed acquired and the distance travelled in this time.​

Answer 1

Answer:

Explanation:

Correct Question,

(Time taken is not given)

A motor cycle moving with speed of 5m/s is subject to an acceleration of 0.2m/s²

. Calculate the speed of motorcycle after 10 second, and the distance travelled in this time.

Answer,

Given,

Initial velocity, u = 5 m/s

Acceleration, a = 0.2 m/s²

Time taken, t = 10 seconds

To Find,

Final velocity, v = ?

Distance covered, s = ?

Formula to be used,

1st equation of motion, v = u + at

3rd equation of motion, v² - u² = 2as

Solution,

Putting all the values, we get

v = u + at

⇒ v = 5 + 0.2 × 10

⇒ v = 5 + 2

⇒ v = 7 m/s

Hence, the speed acquired is 7 m/s.

Now, the distance covered,

v² - u² = 2as

⇒ (7)² - (5)² = 2 × 0.2 × s

⇒ 49 - 25 = 0.4s

⇒ 24 = 0.4s

⇒ 24/0.4 = s

⇒ s = 60 m

Hence, the distance covered is 60 m.

Answer 2

Answer:

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

Distance traveled in 10 seconds = 60 m

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/s

Hope it helps u !!