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A vehicle is accelerating on a straight road. Its velocity at any instant is 30 km h-1, after 2 s, it is 33.6 km h-1 and after further 2 s, it is 37.2 km h-1. Find the acceleration of vehicle in m s-2. Is the acceleration uniform ?

Motion in One Dimension

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Answer

As we know,

a = v−ut\dfrac{v - u}{t}

After t = 2s,

u = 30 km h-1

Converting km h-1 to m s-1

We get,

30km h−1=30km1h=30×1000m60×60s⇒30km h−1=30×10m6×6s⇒30km h−1=8.33×ms30 \text{km h}^{-1} = \dfrac{30 \text{km}}{1\text{h}} = \dfrac {30 \times 1000 \text{m}}{60 \times 60 \text{s}} \\[0.5em] \Rightarrow 30 \text{km h}^{-1} = \dfrac {30 \times 10 \text{m}}{6 \times 6 \text{s}} \\[0.5em] \Rightarrow 30 \text{km h}^{-1} = \dfrac {8.33 \times \text{m}}{\text{s}} \\[0.5em]

Hence, 30 km h-1 is equal to 8.33 m s-1.

Now,

v = 33.6 km h-1

Converting km h-1 to m s-1

We get,

33.6km h−1=33.6km1h=33.6×1000m60×60s⇒33.6km h−1=33.6×10m6×6s⇒33.6km h−1=9.33×ms33.6 \text{km h}^{-1} = \dfrac{33.6 \text{km}}{1\text{h}} = \dfrac {33.6 \times 1000 \text{m}}{60 \times 60 \text{s}} \\[0.5em] \Rightarrow 33.6 \text{km h}^{-1} = \dfrac {33.6 \times 10 \text{m}}{6 \times 6 \text{s}} \\[0.5em] \Rightarrow 33.6 \text{km h}^{-1} = \dfrac {9.33 \times \text{m}}{\text{s}} \\[0.5em]

Hence, 33.6 km h-1 is equal to 9.33 ms-1.

Substituting the values in the formula above, we get,

a=9.33−8.332a=12⇒a=0.5ms2a = \dfrac{9.33 - 8.33}{2} \\[0.5em] a = \dfrac{1}{2} \\[0.5em] \Rightarrow a = 0.5 m s^{2} \\[0.5em]

Hence, acceleration in the first 2 s = 0.5 m s -2

For the next 2 s,

u = 9.33 m s-1

v = 37.2 km h-1

Converting km h-1 to m s-1

We get,

37.2km h−1=37.2km1h=37.2×1000m60×60s⇒37.2km h−1=37.2×10m6×6s⇒37.2km h−1=10.33×ms37.2 \text{km h}^{-1} = \dfrac{37.2\text{km}}{1\text{h}} = \dfrac {37.2 \times 1000 \text{m}}{60 \times 60 \text{s}} \\[0.5em] \Rightarrow 37.2 \text{km h}^{-1} = \dfrac {37.2 \times 10 \text{m}}{6 \times 6 \text{s}} \\[0.5em] \Rightarrow 37.2 \text{km h}^{-1} = \dfrac {10.33 \times \text{m}}{\text{s}} \\[0.5em]

Hence, 37.2 km h-1 is equal to 10.33 ms-1.

Substituting the values in the formula above, we get,

a=10.33−9.332a=12⇒a=0.5ms2a = \dfrac{10.33 - 9.33 }{2} \\[0.5em] a = \dfrac{1}{2} \\[0.5em] \Rightarrow a = 0.5 m s^{2} \\[0.5em]

Hence, acceleration is 0.5 m s -2

Yes, the acceleration is uniform as the acceleration in both instances is same.

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