Question

I am stuck on these questions below, thanks in advance for your help!

An object is moving with constant non-zero velocity in the +x axis. The position versus time graph of this object is a parabolic curve. a hyperbolic curve. a vertical straight line. a straight line making an angle with the time axis. a horizontal straight line. Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 2.40 m/s2. How long does It take car A to overtake car B? 19.0 s 12.6 s 316 s 5.50 s Car A never overtakes car B. If a seagull drops a shell from rest at a height of 18 m, how fast is the shell moving when it hits the rocks? Use g=9.81 m/s 2 18 m/s 19 m/s 20 m/s 21 m/s 25 m/s Suppose that an object travels from one point in space to another. Make a comparison between the displacement and the distance traveled. The displacement is cither less than or equal to the distance traveled. If the displacement is equal to zero, then the distance traveled will also equal zero. The displacement is cither greater than or equal to the distance traveled. The displacement can be cither greater than, smaller than, or equal to the distance traveled. The displacement is always equal to the distance traveled. A car is traveling at 25 m/s when it runs off the road and hits a utility pole. The car stops instantly, but the driver continues to move forward at 25 m/s. The airbag starts from rest with a constant acceleration from a distance of 50 cm away from the driver and makes contact with him in 9 ms. What is the acceleration of the airbag? 680 m/s2 68000 m/s2 370000 m/s2 6800 m/s2 68 m/s2

Answer 1

1) X = V*t

it is a equation of straight line

answer = D

2) equation of motion for car A

Sa = Ua t + 0.5a t^2

Sb = Ub t

to over take B

Sa - Sb = 300

-7t +.5*2.4*t^2 = 300

t = 18.994 sec

answer = A

3) V^2 -u^2 = 2a h

u = 0

V = sqrt ( 2*g*h)

V = 18.79

answer = B

4) displacement is the shortest distance of two points

there fore

answer is = A

5) distance travelled by air bag

D = 0.5-( 25*9*10^-3)

D = 0.275m

D = 0.5 * a * t* t

a = 6790.123 m/ sec^2

answer = D

Answer 2

1) D

2) A

3) B

4) A

5) D