Question

Answer the following questions:
1:
A car, moving with speed 108 km/h on a straight road, decreases its speed uniformly to 90 km/h in 5 seconds. The displacement of the car in the last second of its motion, before it comes to a halt, is

2:
A ball of mass m is dropped from height h. The time taken by the ball to reach the ground is [Take g as acceleration due to gravity]

Answer 1

1. initial velocity, u = 108 km/h = 108 × 5/18 = 30 m/s

final velocity, v = 90 km/h = 90 × 5/18 = 25 m/s

time taken, t = 5s

using formula, v = u + at

so, 25 = 30 + a(5)

⇒a = -1 m/s²

time taken to come in rest , t = (v - u)/a

= (0 - 30)/-1 = 30 sec

distance travelled in 30th second , S = 30 + (-1)/2 (30 × 2 - 1)

= 30 - 30 + 1/2

= 0.5 m

2. initial velocity of ball, u = 0

using formula, s = ut + 1/2 at²

here, s = h , u = 0, a = g

so, h = 0 + 1/2 gt²

⇒t = √{2h/g}

Answer 2

Answer:

1. initial velocity, u = 108 km/h = 108 × 5/18 = 30 m/s

final velocity, v = 90 km/h = 90 × 5/18 = 25 m/s

time taken, t = 5s

using formula, v = u + at

so, 25 = 30 + a(5)

⇒a = -1 m/s²

time taken to come in rest , t = (v - u)/a

= (0 - 30)/-1 = 30 sec

distance travelled in 30th second , S = 30 + (-1)/2 (30 × 2 - 1)

= 30 - 30 + 1/2

= 0.5 m

2. initial velocity of ball, u = 0

using formula, s = ut + 1/2 at²

here, s = h , u = 0, a = g

so, h = 0 + 1/2 gt²

⇒t = √{2h/g}

please dont forget these sums yeah ok.