Motion in One Dimension — Question 3

As we know,

a = $\dfrac{\text v - \text u}{\text t}$

After t = 2s,

u = 30 km h^-1

Converting km h^-1 to m s^-1

We get,

$30 \text{km h}^{-1} = \dfrac{30 \text{km}}{1\text{h}} = \dfrac {30 \times 1000 \text{m}}{60 \times 60 \text{s}} \\[0.5em] \Rightarrow 30 \text{km h}^{-1} = \dfrac {30 \times 10 \text{m}}{6 \times 6 \text{s}} \\[0.5em] \Rightarrow 30 \text{km h}^{-1} = \dfrac {8.33 \times \text{m}}{\text{s}} \\[0.5em]$

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,

$33.6 \text{km h}^{-1} = \dfrac{33.6 \text{km}}{1\text{h}} = \dfrac {33.6 \times 1000 \text{m}}{60 \times 60 \text{s}} \\[0.5em] \Rightarrow 33.6 \text{km h}^{-1} = \dfrac {33.6 \times 10 \text{m}}{6 \times 6 \text{s}} \\[0.5em] \Rightarrow 33.6 \text{km h}^{-1} = \dfrac {9.33 \times \text{m}}{\text{s}} \\[0.5em]$

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

Substituting the values in the formula above, we get,

$\text a = \dfrac{9.33 - 8.33}{2} \\[0.5em] \text a = \dfrac{1}{2} \\[0.5em] \Rightarrow a = 0.5\ \text {m s}^{2} \\[0.5em]$

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,

$37.2 \text{km h}^{-1} = \dfrac{37.2\text{km}}{1\text{h}} = \dfrac {37.2 \times 1000 \text{m}}{60 \times 60 \text{s}} \\[0.5em] \Rightarrow 37.2 \text{km h}^{-1} = \dfrac {37.2 \times 10 \text{m}}{6 \times 6 \text{s}} \\[0.5em] \Rightarrow 37.2 \text{km h}^{-1} = \dfrac {10.33 \times \text{m}}{\text{s}} \\[0.5em]$

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

Substituting the values in the formula above, we get,

$\text a = \dfrac{10.33 - 9.33 }{2} \\[0.5em] \text a = \dfrac{1}{2} \\[0.5em] \Rightarrow \text a = 0.5\ \text {m s}^{2} \\[0.5em]$

Hence, acceleration is 0.5 m s ^-2

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