Question

numericals :

1. a body starts from rest with a uniform acceleration of 2 m s⁻². find the distance covered by the body in 2 s.
2. a body starts with an initial velocity of 10 m s⁻¹ and acceleration 5 m s⁻². find the distance covered by it in 5 s.
3. a vehicle is accelerating on a straight road. its velocity at any instant is 30 km h⁻¹, after 2 s it is 33.6 km h⁻¹ and after further 2 s, it is 37.2 km h⁻¹. find the acceleration of vehicle in m s⁻². is the acceleration uniform ?

Explanation:

Problem 1:

Step1: Identify the formula for distance in uniform acceleration from rest.

The formula for distance \( s \) covered by a body starting from rest (\( u = 0 \)) with uniform acceleration \( a \) in time \( t \) is \( s=ut+\frac{1}{2}at^{2} \). Since \( u = 0 \), the formula simplifies to \( s=\frac{1}{2}at^{2} \).

Step2: Substitute the given values.

Given \( a = 2\space m/s^{2} \) and \( t = 2\space s \). Substitute these values into the formula: \( s=\frac{1}{2}\times2\times(2)^{2} \).

Step3: Calculate the distance.

First, calculate \( (2)^{2}=4 \). Then, \( \frac{1}{2}\times2 = 1 \). Multiply by 4: \( s = 1\times4=4\space m \).

Step1: Identify the formula for distance in non - zero initial velocity and uniform acceleration.

The formula for distance \( s \) covered by a body with initial velocity \( u \), acceleration \( a \) in time \( t \) is \( s = ut+\frac{1}{2}at^{2} \).

Step2: Substitute the given values.

Given \( u = 10\space m/s \), \( a = 5\space m/s^{2} \) and \( t = 5\space s \). Substitute into the formula: \( s=10\times5+\frac{1}{2}\times5\times(5)^{2} \).

Step3: Calculate each term.

First term: \( 10\times5 = 50 \). Second term: \( \frac{1}{2}\times5\times25=\frac{125}{2}=62.5 \).

Step4: Sum the terms.

\( s=50 + 62.5=112.5\space m \) (Note: There might be a mistake in the provided answer key as per the given data, but following the calculation steps)

Step1: Convert the velocities from km/h to m/s.

To convert from km/h to m/s, use the conversion factor \( 1\space km/h=\frac{1000}{3600}=\frac{5}{18}\space m/s \).

• Initial velocity \( u = 30\space km/h=30\times\frac{5}{18}=\frac{150}{18}=\frac{25}{3}\space m/s\approx8.33\space m/s \)
• Velocity after 2 s \( v_1 = 33.6\space km/h=33.6\times\frac{5}{18}=\frac{168}{18}=\frac{28}{3}\space m/s\approx9.33\space m/s \)
• Velocity after 4 s (2 s + 2 s) \( v_2 = 37.2\space km/h=37.2\times\frac{5}{18}=\frac{186}{18}=\frac{31}{3}\space m/s\approx10.33\space m/s \)

Step2: Calculate acceleration using the formula \( a=\frac{v - u}{t} \) for the first 2 - second interval.

For the first 2 - second interval, \( u=\frac{25}{3}\space m/s \), \( v=\frac{28}{3}\space m/s \), \( t = 2\space s \).
\( a=\frac{\frac{28}{3}-\frac{25}{3}}{2}=\frac{\frac{3}{3}}{2}=\frac{1}{2}=0.5\space m/s^{2} \)
We can also check with the second 2 - second interval. For the second 2 - second interval, \( u=\frac{28}{3}\space m/s \), \( v=\frac{31}{3}\space m/s \), \( t = 2\space s \).
\( a=\frac{\frac{31}{3}-\frac{28}{3}}{2}=\frac{\frac{3}{3}}{2}=\frac{1}{2}=0.5\space m/s^{2} \)

Answer:

The distance covered by the body is \( \boldsymbol{4\space m} \).

Problem 2: