Question

A moter cycle moving with a speed of 5m/s obtain to an acceleration of 0.2 m/s sq . calculate the speed of the moter cycle after 10 s and the distance travelled by it in this time

Answer 1

Answer:

s= distance = 60 m

speed = 7 m/s

Explanation:

Initial velocity = u = 5m/s

Acceleration = a = 0.2m/s²

Time = t = 10 seconds

To find:

• Speed after 10 seconds
• Distance after 10 seconds

Using the first equation of motion,i.e. v=u+at, we can find the speed after 10 seconds

substituting the values,

v=5+(0.2×10)

v=5+2

v=7m/s

The distance can be found with the second equation of motion, i.e. $s=ut+\frac{1}{2}at^{2}$

substituting the values,

$s=5\times 10+\frac{1}{2}\times0.2\times10^{2}$

$s=50+10$

$s=60m$

s=60m

Answer 2

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QUESTION :

A moter cycle moving with a speed of 5m/s obtain to an acceleration of 0.2 m/s sq .

Calculate the speed of the moter cycle after 10 s and the distance travelled by it in this time.

SOLUTION :

In the above Question, the following information is given.

U = 5 m / s

A = 0.2 m / s

1. calculate the speed of the moter cycle after 10 s.

Let after 10 seconds, the speed be V

According to Newton ' s first law of motion ,

=> V - 5 / 10 = 0.2

=> V = 7 m / s

So, the answer of the first part is 7 m / s.

2. the distance travelled by it in this time.

According to Newton's Second law of motion ,

S = ut + { 1 / 2 } a t ^ 2

Here, t = 10 s.

U = 5 m / s

A = 0.2 m / s ^ 2

Substituting the above values we get :

S = 5 × 10 + { 1 / 2 } × 0.2 × 10 ^ 2

=> S = 50 + 10

=> S = 60 m.

ANSWER :

The speed of the moter cycle after 10 s is 7 m / s and the distance travelled by it in this time is 60 m.