Question

A car accelerates uniformly from 18 km h-¹ to 36 I'm h-¹ in 5s. Calculate:- (I) the Acceleration (II) Distance covered by the car in that time.

Answer 1

Given

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

To find

1. Acceleration
2. Distance travelled by car in that time

Solution

Here we get the values as ;

Initial speed (u) = 18 km/h = (18 × 5/18) m/s = 5 m/s

Final speed (v) = 36 km/h = (36 × 5/18) m/s = 10 m/s

We will use 1st equation of motion to find acceleration :

⟼ v = u + at

Substituting values,

➺ 10 = 5 + a × 5

➺ 10 - 5 = 5a

➺ 5 = 5a

➺ a = 5/5

➺ a = 1 m/s²

∴ Acceleration of car = 1 m/s²

Now we will use 3rd equation of motion to find distance travelled by car :

⟼ v² - u² = 2as

Substituting values,

➺ 10² - 5² = 2 × 1 × s

➺ 100 - 25 = 2s

➺ 75 = 2s

➺ s = 75/2

➺ s = 37.5 m

∴ Distance travelled by car = 37.5 m

Answer 2

Answer:

Acceleration = Change in Velocity/Time

Change in Velocity = 36-18 = 18 km/h=5 m/s

Time= 5 Seconds

Acceleration = 5/5= 1 m/s2

Equation of motion,s=ut+(1/2)at2

u=18 km/h=5 m/s

t=5 s

a=1 m/s2

s= (5*5)+(1/2*1*5*5)

s=25+12.5 i.e., s=37.5 m

Hope you are clear with my explanations