Write your solutions in the form of variables first, and plug in numbers at the

Question

Write your solutions in the form of variables first, and plug in numbers at the last step. A car at rest starts at the origin and accelerates at 2m/s^2 in the +y direction for 5 seconds, and continues to coast at what whatever speed it reaches in the y direction. At 5 seconds, it also starts to accelerate at 5m/s^2 for 3 seconds in the +x direction. a. Calculate the y-position of the car before it starts to accelerate to the +x direction. 25m b. Calculate the final x-position of the car. 22.5m c. Calculate the final position of the car. (22.5m, 55m) d. Draw the path that the car travels. Not all lines will be straight! e. Graph the y position vs time. f. Graph the x-position vs time.

Explanation / Answer

a] y = uyt + (1/2)ayt2

=> y = 0 + (1/2)(2)(5)2 = 25 meters

b] Displacement in x direction =

x = uxt + (1/2)axt'2 = 0 + (1/2)(5)(3)2 = 45/2 = 22.5 meters

c]

After 5 seconds, the car continues to travel in y direction but with constant velocity. It also has an acceleration of 5m/s2 in x direction now.

so velocity in y direction after 5 seconds is: vy = uy + ayt = 0 + 2(5) = 10 m/s

therefore, total displacement in y direction will be: Y = y + vt = 25 + 10(3) = 55 meters

=> Final position of the car will be: D = (x,y) = (22.5, 55) meters.

d] After 5 seconds, the y increases linearly (since no acceleration in y direction) and x increases quadratically (since acceleration in x direction is non zero now) with time. The relation between x and y displacement after 5 seconds will be:

x = (1/2)ay(y/vy)2 = 0.025y2

=> y = 6.324x1/2

So, the path until the first 5 seconds will be vertically upwards and then after 5 seconds, the path will be curving parabolically [Similar to y = x1/2 graph].

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