Question

a car acceleration uniformly form 15 km/h to 36 km/h in 5 sec calculate 1)find the acceleration and 2) distance cover by this time

Answer 1

Given:-

• Initial velocity of car (u) = 15km/h = 15 ×5/18 = 75/18 m/s.

• Final velocity of car (v) = 36km/h = 36 ×5/18 = 10m/s

• Time taken (t) = 5 sec.

To Find:-

• Acceleration of the car (a) .
• Distance covered by car (s) .

Solution:-

By using 1st equation of motion

⟹ v = u+at

Putting all the values which is given above

⟹ 10 = 75/18 + a×5

⟹ 10- 75/18 = a×5

⟹ 180-75/18 = a ×5

⟹ 105/18 = a ×5

⟹ a = 105/18×5

⟹ a = 21/18

⟹ a = 7/6 m/s² or 1.16m/s²

∴ Acceleration of the car is 1.16m/s²

And Now ,

By using 2nd equation of motion

⟹ s = ut +1/2at²

Putting all the values which is given above

⟹ s = (75/18) ×5 + 1/2 × 1.16 × 5²

⟹ s = 375/18 + 0.58 ×25

⟹ s = 20.83 + 14.5

⟹ s = 35.33m

∴ The Distance covered by the car at that time is 35.33m