Question

Find the time taken for a car that starts from rest and accelerates at 2m/s2 over a distance of 200m. show your working.

Answer 1

Answer:

Answer given in photo. Follow me.

Answer 2

Answer- The above question is from the chapter 'Kinematics'.

Some important terms and formulae:-

1. Velocity- It is the displacement per unit time.

S.I. Unit of Velocity- m/s

It is a vector quantity as it possesses magnitude and direction.

2. Acceleration- It is the rate of change of velocity.

S.I. Unit of Acceleration- m/s²

It is also a vector quantity.

Negative acceleration is called retardation.

3. Distance- It is the path length transversed by an object.

S.I. Unit of Distance- m

It is a scalar quantity.

4. Displacement- It is the shortest distance between the initial and final point.

S.I. Unit of Displacement- m

It is a vector quantity.

5. Equations for uniformly accelerated motion-

Let u = Initial velocity of a particle

v = Final velocity of a particle

t = Time taken

s = Distance travelled in the given time

a = Acceleration

1) v = u + at

2) s = $\frac{1}{2}$ at² + ut

3) v² - u² = 2as

6. Average Speed = Total distance ÷ Total time

7. Average Velocity = Total displacement ÷ Total time

Given question: Find the time taken for a car that starts from rest and accelerates at 4 m/s² over a distance of 200 m. Show your working.

(Please correct the question.)

Answer: Initial velocity (u) = 0 m/s

Acceleration (a) = 4 m/s²

Distance (s) = 200 m

Time taken (t) = ?

Using 2nd equation of motion, s = $\frac{1}{2}$ at² + ut

200 = [tex] \frac{1}{2} × 4 × t² + 0 × t

200 ÷ 2 = t²

t = √100

t = 10 s

∴ Time taken by the car = 10 seconds.