Question

Solve the following problems. (follow the 5 steps)
1. An object accelerates from rest to a velocity of 4.5 m/s in 3 seconds. Solve for the object’s acceleration. 2. A ball was tossed upwards at a velocity of 30 m/s. What is its velocity after 2 seconds if it has an acceleration of –9.8 m/s²?
3. A car is traveling with an initial velocity of 12 m/s. How long will it take before the car stops if it has an acceleration of –1.2 m/s²?

Answer 1

Answer:

1.An object accelerates from rest to a velocity 27.5 m/s in 10 second.

Given that, the initial velocity of the object is 0 m/s (as it starts from rest), final velocity is 27.5 m/s and time is 10 seconds.

We have to find the distance covered by object in next 10 second.

Using the First Equation Of Motion,

v = u + at

Substitute the known in above formula,

→ 27.5 = 0 + a(10)

→ 27.5 = 10a

→ 27.5/10 = a

→ 2.75 = a

Therefore, the acceleration of the object is 2.75 m/s².

Now,

Using the Second Equation Of Motion,

s = ut + 1/2 at²

In next 10 seconds, initial velocity becomes 27.5 m/s.

Substitute the known in the above formula,

→ s = 27.5(10) + 1/2 × 2.75 × (10)²

→ s = 275 + 1/2 × 2.75 × 100

→ s = 275 + 1/2 × 275

→ s = 275 + 137.5

→ s = 412.5

Therefore, the distance covered by the object in next 10 second is 412.5 m.

2.Initial velocity 30m/s

If velocity after 2sec =v , where f = -9.8m/sec^2

v= u+ft = 30+(-9.8)*2=30–19.6 = 10.4

After 2 sec velocity will be 10.4 m/sec

If the ball is at a hight h

Then h = ut +0.5 ft^2= 30*2+0.5(-9.8)(2)^2

Or, h = 60–19.6 = 40.4 m

Position of the ball is at a hight 40.4 m

3.v = u + at

0 = 12 + (3.5) × a

a = - (12) / (3.5)

a = - 3.43 m/s2

Explanation: