Question

Example 8.6 A car accelerates uniformly
from 18 km h to 36 km h in 5 s.
Calculate (i) the acceleration and (ii) the
distance covered by the car in that time.​

Answer 1

⭐ Given :-

• Initial Velocity(U) = 18km/h = 5m/s

• Final Velocity(V) = 36km/h = 10m/s

• Time = 5 seconds.

⭐ To Find :-

• Acceleration and Distance Covered.

⭐ Solution :-

Acceleration = (V - U)/T

➣ Acceleration = (10 - 5)/5

➣ Acceleration = 5/5

➣ Acceleration = 1m/s².

Hence, Acceleration = 1m/s².

Now, Distance :-

Use Second Equation Of Motion.

s = ut + ½ at²

Put the values.

➣ s = 5×5 + ½ × 1 × 5 × 5.

➣ s = 25 + 25/2

➣ s = 25 + 12.5

➣ s = 37.5m

Hence, The Distance Traveled = 37.5 m.

Answer 2

Step-by-step explanation:

v = 36km/h = 36×5/18

= 10m/s

u= 18 km /h = 18 × 5/ 18

= 5m/s

v= u +at ............(1)

10= 5 + a× 5

10- 5= 5a

5= 5a

5/5= a

1 =a ( acceleration is 1 m/s²)

distance (s)= ut + 1/2 at² ................(2)

s = 5×5 + 1/2 × 5× 5

s =25+ 12.5

s = 37.5 m