Question

A motor cycle attains a velocity of 72 kmh-1 in 20 seconds just after the start.
Calculate the acceleration of the motor cycle

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Acceleration\:of\:motor\:cycle=1\:m/s^{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies Final \: velocity(v) = 72 \: km/h \\ \\ \tt: \implies Time(t) = 20 \: sec \\ \\ \red{\underline \bold{To \: Find:}}\\ \tt: \implies Acceleration(a) = ?$

• According to given question :

$\tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \tt \circ \: Final \: velocity =72 \times \frac{5}{18} = 20 \: m/s \\ \\ \tt \circ \: time = 20 \: sec \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 20 = 0 + a \times 20 \\ \\ \tt: \implies 20 = 20 \times a \\ \\ \tt: \implies a = \frac{20}{20} \\ \\ \green{\tt: \implies a = 1 \: {m/s}^{2} } \\ \\ \green{\tt \therefore Acceleration \: of \: motor \: cycle \: is \: 1 \: {m/s}^{2} } \\ \\ \blue{ \bold{Some \: related \: formula}} \\ \orange{\tt \circ \: {v}^{2} = {u}^{2} + 2as} \\ \\ \orange{\tt \circ \: s = ut + \frac{1}{2} a {t}^{2} }$

Answer 2

Answer:

Given:

• A motor cycle attains a velocity of 72 km/h-1 in 20 seconds just after the start.

Find:

• Calculate the acceleration of the motor cycle.

According to the question:

• 0 m/s = Initial velocity.

• 20 m/s = Final velocity.

• 20 seconds = Time.

Know terms:

• (v) = Velocity.

• (u) = Initial velocity.

• (a) = Acceleration.

• (t) = Time.

Using formula:

⇒ v = u + at.

Calculations:

⇒ 20 = 0 + a × 20

⇒ 20 = 20 × a

⇒ a = 20/20

⇒ a = 1 m/s²

Therefore, 1 m/s² is the acceleration of the motor cycle.