PHYSICS - A foreign automobile manufacturer claims that its sports car will accelerate from rest to a speed

Question

A foreign automobile manufacturer claims that its sports car will accelerate from rest to a speed of 45.5 m/s in 7.35 s.

(a)

Determine the magnitude of the average acceleration of the car (in m/s²).

(b)

Assume that the car moves with constant acceleration. Find the distance (in m) the car travels in the first 7.35 s.

 

(c)

What is the speed of the car (in m/s) 10.0 s after it begins its motion if it continues to move with the same acceleration?

 

Answer 1

a) use the equation Vf=Vi+at

Vf=45.5 m/s

Vi=0 m/s (at rest)

a=?

t=7.35 sec

distance is unimportant to solving letter a.

plug in:

45.5=0+a(7.35)

a=45.5/7.35=6.19 m/s2

b) use the equation d=Vixt+(1/2)at^2 or Vf^2=Vi^2+2ad

Vf=45.5 m/s

Vi=0 m/s (at rest)

a=6.19 m/s2

t=7.35 sec

d=?

plug in:

d=(1/2)(6.19)(7.35)^2

d=167.20 m

c) use the equation Vf= Vi+at

Vf= ? m/s

Vi=0 m/s (at rest)

a=6.19 m/s2

t= 10 sec

d= unimportant and unknown

Vf=(6.19)(10)= 61.9 m/s

Answer 2

acceleration=(v-vo)/t=45.5m/s/7.35s=6.49m/s^2

constant=45.5*7.35=334.425m

speed-10*6.49=64.9m/s