Question

on spotting a prey, a cheetah runs directly towards it with constant acceleration. The time taken by the cheetah is 50s and it's velocity, as it catches it's prey , is 25m/s.If we assume that the cheetah was initially at rest , then what is it's acceleration​

Answer 1

Answer:

acceleration is 0.5 m/s²

Step-by-step explanation:

u= 0 m/s

v= 25 m/s

t= 50 s

a= (v-u)/t

= (25-0)/50

= 25/50

= 1/2

= 0.5 m/s²

Answer 2

Answer:

0.5 m/s²

Step-by-step explanation:

Given:

• Initial velocity, u = 0 m/s. ( as we have assumed that the cheetah was initially at rest, so initial velocity is zero )
• Final velocity, v = 25 m/s
• Time taken, t = 50 s

To find:

• Acceleration, a.

Solution:

Using first equation of motion. Putting the values in the equation and then doing required calculations, we will find the value of a a which is the constant acceleration of cheetah.

Using first equation of motion,

• v = u + at

Where,

v is the final velocity.

u is the initial velocity.

a is the acceleration.

t is the time taken.

Here,

v = 25 m/s

u = 0 m/s

a = ? ( we need to find )

t = 50 s

Putting the values,

• 25 = 0 + a × 50
• 25 = 0 + 50a
• 25 = 50a
• a = 25/50
• a = 5/10
• a = 0.5 m/s²

∴ The constant acceleration of cheetah = 0.5 m/s²

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