A0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters and t in seconds. at t = 1.1 s, what are (a) the magnitude and (b) the angle (within (-180°, 180°] interval relative to the positive direction of the x axis) of the net force on the particle, and (c) what is the angle of the particle's direction of travel?

Part a)

Part b)

Part c)

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

Part a)

differentiate x and y two times with respect to time to find the acceleration

Now the acceleration of the object is given as

at t= 1.1 s we have

now the net force of the object is given as

now magnitude of the force will be

Part b)

Direction of the force is given as

Part c)

For velocity of the particle we have

now at t = 1.1 s

now the direction of the velocity is given as

7.21 ft/s^2
       
Since you're looking for average acceleration, you can simply divide the change in velocity by the time. To make the calculation more reasonable, first convert the speed of 173 mi/h into ft/sec by multiplying by 5280 to convert from mi/h to ft/h and then dividing by 3600 to convert from ft/h to ft/s. 173 * 5280 / 3600 = 253.7333 ft/s Now divide the change in velocity by the time in seconds. 253.7333 ft/s / 35.2 s = 7.208333 ft/s^2 Rounding result to 3 significant figures gives 7.21 ft/s^2