Question

The mass of a car is 1400 kg. The car, initially at rest, is moved along a level road by a resultant

force of 3500 N. The car reaches a speed of 30 m / s
(a) Calculate the average acceleration of the car.
(b) Calculate the time for which the force is applied.
(c) State the name of a force which opposes the motion of the car.​

Answer 1

Answer:

i) a=30m/sec²

ii) 1 sec

iii) Frictional Force/Friction

Explanation:

M=1400kg

u=0

v=30m/Sec

F=3500Nt=1s

i) a = v-u/t

a=30-0/1

=a=30m/sec²

ii) 1 sec

iii) Friction

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Answer 2

Given :

Mass of the car = 1400kg

Initial velocity = zero

Resultant force = 3500N

Finql velocity = 30m/s

To Find :

• Acceleration of car
• Time of journey
• Name of a force which opposes the motion of car

Solution :

❖ As per newton's second law of motion, force is defined as the product of mass and acceleration.

➙ F = m a

• F demotes force
• m denotes mass
• a denotes acceleration

➙ 3500 = 1400 × a

➙ a = 3500/1400

➙ a = 2.5 m/s²

♦ Time of journey :

We know that acceleration is defined as the rate of change of velocity$.$

Mathematically, a = (v - u) / t

• v denotes final velocity
• u denotes initial velocity
• t denotes time

➙ 2.5 = (30 - 0) / t

➙ t = 30/2.5

➙ t = 12 s

♦ Frictional force opposes the motion of car.