Question

(17) A train has an initial velocity of 24 km hr-!. After 8 s, its velocity is
72 km hr-1. What is the value of the acceleration ? Calculate the distance
travelled in the 8 s.​

Answer 1

Answer:-

• Acceleration $\large\leadsto\boxed{\sf\green{1.67 \: ms^{-2}}}$

• Distance travelled $\large\leadsto\boxed{\sf\green{106.8 \: m}}$

• Given:-

Initial velocity[u] = 24 kmh-¹ → 6.67 ms-¹

Final velocity [v] = 72 kmh-¹ → 20 ms-¹

Time [t] = 8 sec.

• To Find:-

Acceleration [a] = ?

• Solution:-

Firstly,

• Using the 1st equation of motion:-

$\pink{\bigstar}$$\large\boxed{\sf\purple{v = u + at}}$

→ $\sf{20 = 6.67 + \bf{a} \times 8}$

→ $\sf{20 - 6.67 = 8 \bf{a}}$

→ $\sf{8 {\bf{a}} = 13.33}$

→ $\sf{{\bf{a}} = \dfrac{13.33}{8}}$

→ $\sf{{\bf{a}} = 1.66625 \approx 1.67}$

$\implies\boxed{\bf\red{a = 1.67 \: ms^{-2}}}$

Now, to calculate distance,

• Using the 3rd equation of motion:-

$\pink{\bigstar}$$\large\boxed{\sf\purple{s = ut + \dfrac{1}{2} at^2}}$

→ $\sf{6.67 \times 8 + \dfrac{1}{2} \times 1.67 \times {8}^{2}}$

→ $\sf{53.36 + 0.835 \times 64}$

→ $\sf{53.36 + 53.44}$

$\implies\boxed{\bf\red{s = 106.8 \: m}}$

Therefore, the acceleration will be 1.67 ms-² and the distance travelled in 8 seconds will be 106.8 m.