Question

Velocity of a horse is changing from 20 m/s to 4 m/s in 4 second. Acceleration of the horse is *



- 4 m/s^2

16 m/s^2

- 10 m/s^2

24 m/s^2​

Answer 1

Answer:

–4 m/s²

Explanation:

As per the provided information in the given question, we've :

• Velocity of a horse is changing from 20 m/s to 4 m/s in 4 seconds.

From this statement, we can say that the initial velocity (u) of the horse is 20 m/s. The final velocity (v) of the horse is 4 m/s and the time taken (t) is 4 seconds.

We are asked to calculate its acceleration.

We can find the acceleration in two ways,

• Using the acceleration formula
• Using the first equation of motion

Using the acceleration formula :

$\longmapsto\bf{ a = \dfrac{v -u}{t} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\longmapsto\bf{ a = \dfrac{4 \; ms^{-1} - 20 \; ms^{-1} }{4 \; s} }$

$\longmapsto\bf{ a =\cancel{ \dfrac{-16 \; ms^{-1} }{4 \; s}} }$

$\longmapsto\bf{ a = -4 \; ms^{-2} }$

∴ Acceleration of the horse is –4 m/s².

Using the first equation of motion :

$\longmapsto\bf{ v = u + at}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\longmapsto\bf{ 4 \; ms^{-1} = 20 \; ms^{-1} + 4s(a)}$

$\longmapsto\bf{ 4 \; ms^{-1} - 20 \; ms^{-1} =4s(a)}$

$\longmapsto\bf{ -16 \; ms^{-1} =4s(a)}$

$\longmapsto\bf{ \cancel{ \dfrac{-16 \; ms^{-1} }{4 \; s}} = a }$

$\longmapsto\bf{ -4 \; ms^{-2} =a }$

∴ Acceleration of the horse is –4 m/s².

$\rule{200}2$

More Information :

• Acceleration is the rate of change in velocity.
• It is a vector quantity.
• SI unit of acceleration is m/s².
• Negative acceleration is known as retardation or deceleration.