Question

Usha
3) A car starting from rest attains a velocity of 144 km/hr in 10 minutes. Assuming that the
acceleration is uniform. Find i) the acceleration and ii) the distance travelled by the car for attaining
this velocity​

Answer 1

Given:-

• Initial velocity ,u= 0m/s

• Final velocity , v= 144km/h

• Time taken , t = 10min

To Find:-

• (i) Acceleration, a

• (ii) distance travelled ,s

Solution:-

Firstly we convert the unit here

Final velocity = 144×5/18 = 8×5 = 40m/s

Time taken = 10min = 10×60 = 600s

By using 1st equation of motion

→ v = u+at

Substitute the value we get

→ 40 = 0+ a×600

→ 40 = a×600

→ a = 40/600

→ a = 0.06m/s²

∴ The acceleration of the car is 0.06m/s²

Now by using 2nd equation of motion

→ s = ut +1/2at²

Substitute the value we get

→ s = 0×600 +1/2 ×0.06×600²

→ s = 0 + 0.03 × 600×600

→ s = 0.03×360000

→ s = 10800 or 10.8km

∴ The distance covered by the car is 10800 metre .

Answer 2

Answer:

10800m

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