Question

A cheetah is walking at 1.0 m/s when it sees a zebra 25 m away. what acceleration would be required to reach 20.0 m/s in that distance?

Answer 1

Answer:

5m/s

Explanation:

u need to add

Answer 2

$15.2 m/sec^{2}$ would be required to reach 20.0 m/s in that distance.

Given,

Speed of the cheetah=1.0 m/s

Distance between the cheetah and the zebra=25 m.

To find,

the acceleration that would be required to reach 20.0 m/s in that distance.

Solution:

• Speed is equal to the ratio of the distance covered by an object and the time taken by the object to cover that distance.
• $Speed=\frac{Distance}{Time}$.
• Its S.I. unit is m/sec.
• The rate of change in velocity is called acceleration.
• Its S.I. unit is $m/sec^{2}$.
• The equations of motion which are used in such questions are:
• $v=u+at$.
• $s=ut+\frac{1}{2} at^{2}$.
• $v^{2}=u^{2} +2as$.
• where, v-final velocity, u-initial velocity, a-acceleration and s-displacement.

The time the cheetah takes to reach 25 m while running at 20 m/s,

$Speed=\frac{Distance}{Time} \\20=\frac{25}{Time} \\Time=\frac{25}{20} \\ Time=1.25 seconds.$

The cheetah takes 1.25 seconds to reach 25 m while running at 20 m/s.

Its acceleration that would be required to reach 20.0 m/s in that distance,

$v=u+at\\20=1+1.25a\\20-1=1.25a\\19=1.25a\\a=\frac{19}{1.25} \\a=15.2 m/sec^{2} .$

Hence, its acceleration is $15.2 m/sec^{2}$.

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