A car accelerates uniformly from 18 km h⁻¹ to 36 km h⁻¹ in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Class 9th
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A car accelerates uniformly from 18 km h⁻¹ to 36 km h⁻¹ in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Class 9th
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A car accelerates uniformly from 18 km h⁻¹ to 36 km h⁻¹ in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
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Here, We have initial and final velocity and time taken.
Initial velocity, u is given as 18 km/h, So we have to convert it into m/s,
For that we have to multiply it with 5/18,
So, 18 × 5/18 = 5 m/s.
Similarly, the final velocity is given as 36 km/s.
So, 36 × 5/18 = 10 m/s.
Now, time taken is also given as 5 seconds,
Here, we have to find,
Acceleration and Distance covered.
According to the first equation of motion,
We know that,
v = u + at
So, putting all the values, we get
⇒ v = u + at
⇒ 10 = 5 + a × 5
⇒ 10 - 5 = 5a
⇒ 5 = 5a
⇒ 5/5 = a
⇒ a = 1 m/s².
Hence, the acceleration of the car is 1 m/s².
Now, we have to find the distance covered, s
According to the 3rd equation of motion,
we know that,
v² - u² = 2as
So, putting all the values again, we get
⇒ v² - u² = 2as
⇒ (10)² - (5)² = 2 × 1 × s
⇒ 100 - 25 = 2s
⇒ 75 = 2s
⇒ 75/2 = s
⇒ s = 37.5 m.
Hence, the distance covered by car is 37.5 m.
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