Motion 2 - SS1 Physics Past Questions and Answers - page 1

Using the equation v = u + at, we have:

v = 0 + 4 m/s2 X 5 s

v = 20 m/s

Using the equation v = u + at, we have

30 m/s = 10 m/s + 2 m/s2 X t

20 m/s = 2

Using the equation s = ut + 1/2at2, we have:

s = 0 + 1/2 X 2 m/s2 X (8 s)2

s = 1/2 X 2 m/s2 X 64 s2

s = 64 m

Using the equation s = (v + u) X t / 2:

s = (30 m/s + 10 m/s) X 1 s / 2

s = 40 m/s X 1 s / 2

s = 20 m

Using the equation s = ut + 1/2at2, where u is the initial velocity (0 m/s), a is the acceleration (2 m/s2), and t is the time (15 seconds):

s = 0 + 1/2 X 2 m/s2 X (15 s)2

s = 0 + 1/2 X 2 m/s2 X 225 s2

s = 0 + 225 m

s = 225 metres

Therefore, the car will travel 225 metres in 15 seconds.

To find the maximum height reached by the ball, we need to determine the time it takes for the ball to reach its peak and then use that time to calculate the height using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Given:

u = 15 m/s (upwards)

a = -9.8 m/s^2 (acceleration due to gravity)

To find the time taken to reach the peak, we use the equation v = u + at and set v = 0:

0 = 15 m/s - 9.8 m/s2 X t

9.8 m/s2 X t = 15 m/s

t = 15 m/s / 9.8 m/s2

t ≈ 1.53 seconds

Now, to find the maximum height, we use the equation s = ut + 1/2at^2:

s = 15 m/s * 1.53 s + 1/2 * (-9.8 m/s^2) * (1.53 s)^2

s ≈ 11.46 metres

Therefore, the maximum height reached by the ball is approximately 11.46 metres.

To find the magnitude of deceleration, we use the equation a = (v - u) / t, where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time.

Given:

u = 30 m/s

v = 0 m/s

t = 10 s

a = (0 m/s - 30 m/s) / 10 s

a = -30 m/s / 10 s

a = -3 m/s2

Therefore, the magnitude of the deceleration is 3 m/s2.