Question

Cheetahs can accelerate to a speed of 20.5 m/s in 2.45 s and can continue to accelerate to reach a top speed of 29.7 m/s. Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the x direction point in the direction the cheetah runs Express the cheetah's top speed vhop in miles per hour (mi/h). mi/h top0.0185 Starting from a crouched position, how much time taccel does it take a cheetah to reach its top speed and what distance d does it travel in that time? accel d= | 51.27 If a cheetah sees a rabbit 120.0 m away, how much time otal will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest?

Please help! a bit confused. bottom just asks for total

Answer 1

Given that top speed of Cheetah is

V_top = 29.7 m/sec

1 mile = 1609.34 m

1 hr = 3600 sec

So,

V_top = (29.7 m/sec)*(1 mi/1609.34)*(3600 sec/1 hr)

V_top = (29.7*3600/1609.34) mi/hr

V_top = 66.44 mi/hr

Part B.

Using 1st kinematic equation

V = U + a*t

given that when t = 2.45 sec, then V = 20.5 m/sec

U = initial speed = 0 m/sec

a = acceleration = ?

a = (V - U)/t = (20.5 - 0)/2.45 = 8.38 m/sec^2

Now acceleration of cheetah is constant till it reaches top speed So, time taken by cheetah to reach top speed will be

V_top = U + a*t_accel

t_accel = (V_top - U)/a

V_top = 29.7 m/sec = top speed of cheetah

U = initial speed = 0 m/sec

t_accel = (29.7 - 0)/8.38

t_accel = 3.54 sec

Now during this time distance traveled by cheetah will be Using 2nd kinematic equation

d = U*t_accel + (1/2)*a*t_accel^2

Using known values:

d = 0*3.54 + (1/2)*8.38*3.54^2

d = 52.5 m = distance traveled by cheetah to reach top speed

Part C.

Total distance between cheetah and rabbit, D = 120 m

Now distance traveled by Cheetah to reach top speed, d = 52.5 m

time taken by cheetah to reach top speed. t_accel = 3.54 sec

remaining distance, d1 = 120 - 52.5 = 67.5 m

Since after reaching top speed, cheetah's speed will be constant (29.7 m/sec) So, time taken to travel remaining distance will be

time = distance/Speed

t1 = 67.5 m/(29.7 m/sec) = 2.27 sec

Total time taken by cheetah to catch the rabbit will be

t_total = t_accel + t1

t_total = 3.54 + 2.27

t_total = 5.81 sec

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