1. Two pictures are taken of a moving car 3 seconds apart, each with the shutter

Question

1. Two pictures are taken of a moving car 3 seconds apart, each with the shutter open for 1 ms. The car traverses 8 mm during the first exposure, and 20 mm during the second. (a) Find the average acceleration during the 3-s interval, and (b) the distance traversed if the acceleration was constant. (c) Instead, if it’s known only that the car never decelerated—that its velocity never decreased during the interval, what is the minimum and what is the maximum mathematically possible distance, consistent with the same given information? Illustrate with graphs of velocity vs time.

2. An object is thrown upward at 42 m/s at t'0. (Use g ' 10 m/s2 ) (a) Find (i) the time it takes to reach the highest point, (ii) the maximum height. (b) At t ' 2.5 s, find (i) the velocity, (ii) the height, (iii) how many mm higher it moves during the next 1 ms (this requires no additional calculation for 2-digit accuracy). (c) At what time is it moving upward at 27 m/s? How high is it then? (d) At height 68 m, find (i) the speed, and (ii / iii) the times, on the way up/down, each measured from the initial instant. (Use the answer to part (i) in (ii) and (iii) and you won’t have to solve a quadratic equation.) (iv) Check that the highest point is reached midway between these two instants. (e) Draw a graph of v vs t, indicating all of these instants.

Explanation / Answer

2.)

a i)

Vo = initial velocity = 42 m/s

Vf = final velocity at the highest point = 0 m/s

t = time taken

a = acceleration = - g = - 9.8

using the equation

Vf = Vo + at

0 = 42 - 9.8 t

t = 4.3 sec

ii)

Ymax = maximum height

using the equation

Vf2 = Vo2 + 2 a Ymax

02 = 422 + 2 (-9.8) Ymax

Ymax = 90 m

b)

i)

Vo = initial velocity = 42 m/s

Vf = final velocity at 2.5 sec

t = time taken = 2.5 sec

a = acceleration = - g = - 9.8

using the equation

Vf = Vo + at

Vf = 42 - 9.8 (2.5)

Vf  = 17.5 m/s

ii)

Y = height at t = 2.5 sec

Y = Vo t + (0.5) a t2

Y = 42 (2.5) + (0.5) (- 9.8) (2.5)2

Y = 74.4 m

c)

Vo = initial velocity = 42 m/s

Vf = final velocity = 27 m/s

t = time taken = ?

a = acceleration = - g = - 9.8

using the equation

Vf = Vo + at

27 = 42 - 9.8 t

t = 1.53 sec

h = height = Vo t + (0.5) a t2 = (42) (1.53) + (0.5) (-9.8) (1.53)2 = 52.8 m

d)

i)

h = 68 m

Vf = final speed

Vo = initial speed = 42 m/s

using the equation

Vf2 = Vo2 + 2 a h

Vf2 = (42)2 + 2 (-9.8) (68)

Vf = 20.8 m/s

ii)

h = Vo t + (0.5) a t2

68 = 42 t - 4.9 t2

t = 2.2 and 6.4 sec

iii)

midway of 2.2 and 6.4 is 4.3 sec which is the time to reach maximum height

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