Question

this not maths
this is scince (physics)
law of motion ​

Answer 1

AnswEr :

As per the provided information in the given question we have 2 cases. In the first case we have been given the initial velocity, acceleration and , time ; so in the first case we have to find the distance travelled. Now in the second case, we have been provided with initial velocity, time taken and distance travelled ; we have to find the acceleration in the second case.

Formula to apply :

$\bigstar \boxed{\tt { s = ut + \dfrac{1}{2}at^2 }}$

• s denotes distance travelled
• u denotes initial velocity
• t denotes time
• a denotes acceleration

In the first case, we have :

• Initial velocity, u = 5 m/s
• Acceleration, a = 12 m/s²
• Time taken, t = 3 seconds
• Distance, s = ?

Substitute the values in the 2nd equation of motion,

$\longrightarrow \boxed{\tt { s = ut + \dfrac{1}{2}at^2 }} \\ \\ \longrightarrow\tt{s = 5(3) + \dfrac{1}{2} \times 12 \times (3)^2} \\ \\ \longrightarrow\tt{s = 15 + 1 \times 6 \times 9} \\ \\ \longrightarrow\tt{s = 15 + 54} \\ \\ \longrightarrow \underline{\boxed{\tt{s = 69 \; m}}} \; \red{\bigstar}$

∴ Distance travelled in the first case is 69 m.

Now, in the second case, we have :

• Initial velocity, u = 7 m/s
• Time taken, t = 4 seconds
• Distance, s = 40 m
• Acceleration, a = ?

Again substitute the provided values in the second equation of motion.

$\longrightarrow \boxed{\tt { s = ut + \dfrac{1}{2}at^2 }} \\ \\ \longrightarrow\tt{40 = 7(4) + \dfrac{1}{2} \times a \times (4)^2} \\ \\ \longrightarrow\tt{40 = 28 +\dfrac{1}{2} \times a \times 16} \\ \\ \longrightarrow\tt{40 = 28 + 8a} \\ \\ \longrightarrow\tt{40 - 28 = 8a} \\ \\ \longrightarrow\tt{12 = 8a} \\ \\ \longrightarrow\tt{\cancel{\dfrac{12}{8}} = a} \\ \\ \longrightarrow \underline{\boxed{\tt{1.5 \; ms^{-2} = a }}} \; \red{\bigstar}$

∴ Acceleration in the second case is 1.5 m/s².