Question

A car starts from rest and attains velocity of 20 m/s in 20 sec. The distance 1 point
travelled in the car during this period is: *
100 m
200 m
300 m
O 0
400 m​

Answer 1

Answer:

200 metres

Explanation:

Given:

Initial velocity = u = 0 m/s

Final velocity = v = 20 m/s

Time = t = 20 seconds

To find :

Distance travelled (s)

First let us find the acceleration using the first equation of motion:

V=u+at

20=0+a×20

20=20a

a = $\frac{20}{20}$

a = 1 m/s²

The acceleration of the car is equal to 1 m/s²

Now using the third equation of motion:

V²-u²=2as

20²-0²=2×1×s

400-0=2s

400 = 2s

s = $\frac{400}{2}$

s = 200 metres

The distance travelled by the car is equal to 200 metres

Answer 2

GiVeN :-

A car starts from rest and attains velocity of 20 m/s in 20 sec.

☛ Initial velocity, u = 0 m/s

☛ Final velocity, v = 20 m/s

☛ Time, t = 20 s

To DeTeRmInE :-

The distance travelled by the car during this period of 20 s.

AcKnOwLeDgEmEnT :-

✦ v = u + at

✦ v² - u² = 2as

Where,

• v is the final velocity

• u is the initial velocity

• t is the time taken

• a is the acceleration

• s is the distance covered

SoLuTiOn :-

Using the first equation,

20 = 0 + a(20)

⇒20 = 20a

⇒a = 1 m/s²

$\rule{150}2$

Using the second equation,

(20)² - (0)² = 2 × 1 × s

⇒400 - 0 = 2s

⇒2s = 400

⇒s = 200 m

Therefore, the answer is (B) 200 m.