A certain car is capable of accelerating at a uniform rate of 0.93 m/s2. What is the magnitude of the car’s displacement as it accelerates uniformly from a speed of 86 km/h to one of 98 km/h?
Answer in units of m
3 answers
vf^2=vi^2+2ad solve for d.
be certain to change km/hr to m/s...
v = Vi + a t
Vi = 86 km/h * 1000 m/km * 1 h/3600 s
= 23.9 m/s
Vf = 98 km/h *10/36 = 27.2 m/s
hard way:
27.2 = 23.9 + .93 t
so
t = 3.55 s
x = Xi + Vi t + (1/2) a t^2
x-Xi = displacement = 23.9(3.55)+.465(3.55)^2
= 90.7 m
easy way:
average speed = (23.9+27.2)/2 = 25.55
change in speed/t = .93 = 3.3/t
t = 3.55 sure enough
distance = 25.55*3.55 = 90.7 m
Vi = 86 km/h * 1000 m/km * 1 h/3600 s
= 23.9 m/s
Vf = 98 km/h *10/36 = 27.2 m/s
hard way:
27.2 = 23.9 + .93 t
so
t = 3.55 s
x = Xi + Vi t + (1/2) a t^2
x-Xi = displacement = 23.9(3.55)+.465(3.55)^2
= 90.7 m
easy way:
average speed = (23.9+27.2)/2 = 25.55
change in speed/t = .93 = 3.3/t
t = 3.55 sure enough
distance = 25.55*3.55 = 90.7 m
1 answer