Question

A car, moving with speed 108 km/h on a straight road, decreases its speed uniformly to 90 km/h in 5 seconds.

The displacement of the car in the last second of its motion, before it comes to a halt, is​

Answer 1

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Step-by-step explanation:

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Answer 2

Given:-

→ Initial velocity of the car = 108km/h

→ Final velocity of the car = 90km/h

→ Time taken = 5s

To find:-

→ Displacement of the car in the last second of its motion, before it comes to halt.

Solution:-

Firstly, let's convert the initial and final velocity of the car from km/h to m/s.

=> 1km/h = 5/18m/s

=> 108km/h = 108×5/18

=> 30m/s (u)

=> 90km/h = 90×5/18

=> 25m/s (v)

Acceleration of the car :-

=> v = u + at

=> 25 = 30 + a(5)

=> -5 = 5a

=> a = -5/5

=> a = -1m/s²

Now, let's calculate the time taken(t') by the car to come to rest:-

=> 0 = 30 + (-1)t'

=> -30 = -t'

=> t' = 30s

Hence, the car comes to rest in 30 seconds.

So, displacement of the car in the 30th [last] second:-

=> Sₙₜₕ = u + a/2(2n - 1)

=> S₃₀ₜₕ = 30 + (-1/2)(2×30-1)

=> S₃₀ₜₕ = 30 + (-0.5)(59)

=> S₃₀ₜₕ = 30 - 29.5

=> S₃₀ₜₕ = 0.5m

Thus, displacement of the car in the last second of it's motion is 0.5m .