Question

If a car attains a speed of 36 km /h from rest in 12 minutes, what is the distance covered?
Answer is 3.6 km but I need explanation!​

Answer 1

Answer:

3.6 km

Explanation:

u=0 km/h

v=36km/h

t=12 min

36 km/h in m/s

36×5/18=10 m/s

t=12×60=720 sec

By using first equation

v=u+at

10=0+a×720

a=10/720

=1/72 m/s^2

Finding distance by using 3rd equation

v^2-u^2=2as

10^2-0^2=2×1/72×s

100=s/36

s=3600 m=3.6 km

Answer 2

• Initial velocity (u) = 0 km/hr

=> 0 × $\frac{5}{18}$

=> 0 m/s

(Because in starting car was at rest)

• Final velocity (v) = 36 km/hr

=> 36 × $\frac{5}{18}$

=> 10 m/s

• Time = 12 minutes

=> 12 × 60

=> 720 seconds

________________ [ GIVEN ]

• First we have to find the acceleration (a) of the car then the distance (s) covered by the car.

______________________________

We know that ..

» v = u + at

Put the known values in above formula

→ 10 = 0 + a(720)

→ 720a = 10

→ a = $\frac{10}{720}$

→ a = $\frac{1}{72}$ m/s²

______________________________

Now..

» v² - u² = 2as

Put the known values in above formula

→ (10)² - (0)² = 2 × $\frac{1}{72}$ × s

→ 100 = 2 × $\frac{s}{72}$

→ 50 = $\frac{s}{72}$

→ 3600 = s

→ s = 3600 m

→ s = $\frac{3600}{1000}$

→ s = 3.6 km

_____________________________

3.6 km is the distance covered by the car.

_____________ [ ANSWER ]

_____________________________

OR

• Initial velocity (u) = 0 km/hr

• Final velocity (v) = 36 km/hr

• Time = 12 minutes

=> 0.2 hrs.

________________ [ GIVEN ]

• First we have to find the acceleration (a) of the car then the distance (s) covered by the car.

______________________________

We know that ..

» v = u + at

Put the known values in above formula

→ 36 = 0 + a(0.2)

→ a = 180

______________________________

Now..

» v² - u² = 2as

Put the known values in above formula

→ (36)² - (0)² = 2(180)s

→ 1296 = 360s

→ s = 3.6 km

_____________________________

3.6 km is the distance covered by the car.

_____________ [ ANSWER ]

_____________________________