Question

A car accelerates uniformly from 18km /h to 36km/h in 5sec. Calculate a. Acceleration and b. The distance covered by the car in that time

Answer 1

Solution:

Given,

A car accelerates 18 km/h to 36km/h in 5 sec

We need to find

a) acceleration

b) distance

a) sol:-

Using the formula

a = (v - u)/t

Where

• a : acceleration
• v : final velocity
• u : initial velocity
• t : time

→ a = 36 - 18/5

→ a = 18/5

→ a = 3.6 m/s²

b) sol:-

Using the 3rd equation of motion

S = ut + 1/2 at²

Where

• s : distance
• u : initial velocity
• t : time
• a : acceleration

→ S = 18(5) + 1/2 (3.6)(5)

→ S = 90 + 2.3(5)

→ S = 90 + 11.5

→ S = 101.5 km

Hence , acceleration = 3.6m/s² & distance travelled at that time = 101.5 km.

★ Knowledge enhancer

Derivation of formula a = ( v - u )/t

We know that

acceleration = ∆V/t

Where , ∆V : change in velocity = initial velocity - final velocity

a = v - u/t

Hence , derived

Answer 2

Answer:-

Given:-

• Initial velocity (u) = 18 km/h
• Final velocity (v) = 36 km/h
• Time taken (t) = 5 s

To Find:-

1. Acceleration of the car.
2. Distance it covered during the acceleration time period.

___________...

Finding acceleration:-

We know,

a = (v-u)/t

where,

• a = Acceleration,
• v = Final velocity,
• u = Initial velocity &
• t = Time.

∴ a = [(36-18) km h^-1]/(5s)

➵ a = (18 km/h)/5s

➵ a = (18 × 5/18 m/s)/5s

➵ a = 1 m/s² ...(Ans. 1)

Finding distance travelled:-

We know,

s = ut + ½at²

where,

• a = Acceleration,
• s = Distance,
• u = Initial velocity &
• t = Time.

∴ s = (18 km/h × 5s) + ½ (1 m/s²)(5s)²

➵ s = (5 m/s × 5s) + ½ (1 m/s²)(25 s²)

➵ s = 25m × ½ × 25m

➵ s = 25m + 12.5

➵ s = 37.5 m ...(Ans. 2)