An object moving with uniform acceleration has a velocity of 14.0 cm/s in the po

Question

An object moving with uniform acceleration has a velocity of 14.0 cm/s in the positive x direction when its x coordinate is 2.97 cm. If its x coordinate 2.25 s later is -5.00 cm, what is its acceleration? The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of -5.45 m/s^2 for 4.40 s, making straight skid marks 60.2 m long, all the way to the tree. With what speed does the car then strike the tree? 1.69 m/s A particle moves along the x axis. Its position is given by the equation x = 2.1 + 2.7t - 35t^2 with x in meters and t in seconds. (a) Determine its position when it changes direction. 2.62 m (b) Determine its velocity when it returns to the position it had at t = 0 (Indicate the direction of the velocity with the sign of your answer.) -2.7 m/s

Explanation / Answer

displacement = - 5 - 2.97 = 7.97 cm = 0.0797 m

v0 = 0.14 m/s

t = 2.25s

x = v0t + a t^2 /2

- 0.0797 = 0.14 x 2.25 + (a x 2.25^2 / 2)

a = - 0.16 m/s^2 Or - 16 cm/s^2 ........Ans

9. a = - 5.45 m/s^2

t = 4.40 s

d = 60.2 m

d = v0 t + a t^2 /2

60.2 = (v0 x 4.40) - (5.45 x 4.40^2 / 2)

v0 = 25.7 m/s .......Ans

10. x = 2.1 + 2.7t - 3.5t^2

when it changes its direction,

v = dx/dt = 0

dx/dt = 2.7 - 7t = 0

t = 0.386 s

x = 2.1 + (2.7x 0.386) - (3.5 x 0.386^2) = 2.62 m

(b) at t= 0 ,

x = 2.1 m

2.1 + 2.7 t - 3.5t^2 = 2.1

t = 0 Or 0.77 s

v = dx/dt =2.7 - 7t

t = 0.77s

v =2.7 - 7(0.77) = - 2.7 m/s

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