Question

5 questions with ans based on equation of motion

Answer 1

Q. 01. A ball hits a wall horizontally at 6m/s. It rebounds horizontally at 4.4m/s. The ball is in contact with the wall for 0.04s. What is the acceleration of the ball?

sol. Initial velocity(u) = 6.0m/s
Final velocity(v) = -4.4 m/s (because direction of ball has become opposite)
time = 0.04 s
thus, acceleration (a) = (v-u)/t
a= [-4.4 – 6.0]/ 0.04
a= (-10.4)/0.04
after multiplying -10.4/0.04 by 100/100 (to make calculations simpler), we get
a= -1040/4 = -260 m/s2

Q.2 – Calculate the force needed to speed up a car with a rate of 5ms–2, if the mass of the car is 1000 kg.

sol. According to questions:

Acceleration (a) = 5m/s2 and Mass (m) = 1000 kg, therefore, Force (F) =?

We know that, F = m x a

= 1000 kg x 5m/s2

= 5000 kg m/s2

Therefore, required Force = 5000 m/s2 or 5000 N

Q.3 An Object requires the force of 100 N to achieve the acceleration ‘a’. If the mass of the object is 500 kg what will be the value of ‘a’?

sol. According to the question,

Mass (m) = 500 kg, Force (F) = 100N, Acceleration (a) =?

We know that, Force = Mass x Acceleration or F = m x a

∴
100N=500kg×a

⇒ a=100N / 500 kg
100kg ms-2 / 500 kg
= =
0.2m/s−2

Q.4 During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m s–1.

Sol. Given Time Taken (t) = 5 min = 5 x 60s = 300 s
Speed of signal (v) = 3 × 108 m s–1
Distance of spaceship = v x t = 3 × 108 m s–1 x 300 = 9 x 1010 m= 9 x 107 km

Q.5 A racing car has a uniform acceleration of 4 m s–2. What distance will it cover in 10 s after start?
Sol. Given,
Initial velocity of car (u) = 0 m s–1.
acceleration (a) = 4 m s–2.
time taken (t) = 10 s
Distance covered (S) = ?
because , S = ut + ½at2= 0 + 4x100/2 = 400 / 2 = 200 m

Hope this helps u...