Displacement, Velocity, Time

Question

Q. A Cheetah can accelerate from rest to 25 m/s in 6.22s. Assuming constant accel.,

(a) how far does the cheetah run in this time?

(b) After sprinting for just 3.11 s, is the cheetah's speed 12.5 m/s, greater than that or less than that? Explain.

(c) What is the cheetah's average speed for the first 3.11 s of its sprint? For the second 3.11s of its sprint?

(d) Calculate the distance covered by the cheetah in the first 3.11s and the second 3.11s.

I have (a) down as 77.75m and the answers for the rest of the Qs, I only have trouble understanding how to get to them, particularly figuring out avg. speeds during certain time intervals. Thank you.

Answer 1

(a)77.75m

(b) It is greater because the cheetah is not traveling with a uniform velo. and is accelerating.

Using v=u+at this can be proved

(c) 25m/s

(d) 36.125

Answer 2

The easiest way to answer these questions is to use your five velocity, displacement and time equations. they are

where a = acceleration

v = final velocity

u = initial velocity

d = displacement

t = time

v = d/t

a = (v-u)/t

2ad = v^2-u^2

d= ut + 0.5a(t^2)

d= vt - 0.5a(t^2)

Q. a. v=25 u=0 t=6.22 d=?

a= (v-u)/t

=(25-0)/6.22

=25/6.22

=4.02 m/s^2

d=ut+0.5a(t^2)

d= 0x6.22 + 0.5 x 4.02 x 6.22^2

d= 2.01 x 38.69

d=77.76 m

Q. b. v= ? t=3.11 u=0 a=4.02

a=(v-u)/t

at=v-u

v=at+u

v=4.02 x 3.11 + 0

v= 12.5 m/s

Q. c. average is equal to the sum of the velocities divided by the number of velocities

first 3.11 ---> u= 0 v=12.5

Vav= (u + v)/2

=12.5/2

=6.25 m/s

second 3.11 ---> u= 12.5 v=25

Vav=(u+v)/2

=(12.5 +25)/2

=27.5/2

=13.75

Q.d. first 3.11 ---> u=0 v=12.5 a=4.02

2ad=(v^2)-(u^2)

d=((v^2)-(u^2))/2a

d=((12.5^2) - (0^2))/2x4.02

d=(12.5^2)/8.04

d=19.43 m

second 3.11 ---> v=25 u=12.5 a=4.02

d=((v^2)-(u^2))/2a

d=((25^2)-(12.5^2))/2x4.02

d=(625-156.25)/8.04

d=58.3 m

hope this helped