Question

a car decelerated at a constant rate until it stopped such that it traveled 400 m in 5 s, what was the initial velocity of the car? If an object is moving with a constant acceleration such that its average velocity is 6m/s in 4s with an initial veocity of 2 m/s, what is the acceleration of the car? 2. 3. If a ball is dropped from a 100 m building, how long will it take to hit the ground?

Answer 1

Q1.

Using 2nd kinematic equation:

S = U*t + 0.5*a*t^2

Using given value:

400 = U*5 + 0.5*a*5^2

divide above equation by 5

80 = U + 2.5*a

Using 1st kinematic equation

V = U + a*t

0 = U + a*5

a = -U/5

Using this value in above equation

80 = U + 2.5*a

80 = U + 2.5*(-U/5)

80 = U/2

U = 160 m/sec = initial velocity.

2.

Avg Velocity is given by:

Vavg = displacement/time taken

Given that Vavg = 6 m/sec

So displacement = (6 m/sec)*4 sec = 24 m

Now using 2nd kinematic equation:

S = U*t + 0.5*a*t^2

24 = 2*4 + 0.5*a*4^2

8*a = 16

a = 2 m/sec^2 = acceleration

3.

dropped means initial velocity = 0 m/sec

Using 2nd kinematic equation

S = U*t + 0.5*a*t^2

S = height of building = 100 m

U = 0

a = g = 9.8 m/sec^2

So,

100 = 0*t + 0.5*9.8*t^2

t = sqrt (200/9.8)

t = 4.52 sec = time taken to hit the ground

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