A car accelerates uniformly from rest to a speed of 50 mi/h in 12 s. (a) Find the distance the car travels dur

Question

find the distance traveled and constant acceleration

Answer 1

Remember 3 equation for a motion based on Newtion's law:

v = u + at

s = ut + 1/2a(t^2)

v^2 - u^2 = 2as

Where, u = initial velocity
v = final velocity
a = acceleration (-ve for a falling object)
s = distance travelled by moving object

Convert everything into feet and second.

Here, u = 0 mile/sec;
v = 50 miles/hr = 50 x 5280/3600 (note- 1 mile = 5280 ft)
= 73.33 ft/sec
t = 12 sec
s =?

First find acceleration, a.

v = u + at
73.33 = 0 + a*12
a = 73.33/12 = 6.11 ft/sec^2-
ANSWER acceleration = 6.11 ft/sec^2

Now use v^2 - u^2 = 2*a*s

73.33^2 -0 = 2*6.11*s

or s = (73.33^2)/(2*6.11) = 440 ft

ANSWER-distance = 440 ft

Answer 2

15 mph = 22 ft/sec. 50 mph = 220/3 ft/sec.
v = u + at; 50 = 0 + at or a 50/12 m/sec^2 or 220/36.
s = ut + 1/2 a t^2 or s = 1/2 x 220/36 x 144 = 330 feet

Answer 3

Turn 50mph to feet/sec.
Divide that by 12 (seconds) to get the acceleration (in ft/s/s)
Use suitable equation of motion that links distance to initial velocity (zero in this case), acceleration and time.

You should arrive at 440ft.

Answer 4

u = 0 m/s
v = 50 mi/h = 22.35 m/s
t = 12s

v = u + at
a = (v-u)/t = 1.86 m/s^2

v^2 - u^2 = 2as
s = (v^2 - u^2) / 2a = 11164.33 m

Hope this helps.

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