Question

A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answer to the nearest 100th place. Answer in units of ft/s^2. What is the final velocity at this time? Answer in units of ft/s.

Answer 1

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

$34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s$

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2$

Acceleration is 51.94 ft/s²

$v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s$

Final velocity at this time is 257.63 ft/s