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a car initially moving with a uniform velocity of 18 km h⁻¹ comes to a …

Question

a car initially moving with a uniform velocity of 18 km h⁻¹ comes to a stop in 2 s. find the retardation of the car in s.i. units.

a car accelerates at a rate of 5 m s⁻². find the increase in its velocity in 2 s.

a car is moving with a velocity 20 m s⁻¹. the brakes are applied to retard it at a rate of 2 m s⁻². what will be the velocity after 5 s of applying the brakes?

a bicycle initially moving with a velocity

Explanation:

Response
First Sub - Question (Car accelerating)

Step 1: Recall the formula for acceleration

Acceleration \(a=\frac{v - u}{t}\), where \(v\) is final velocity, \(u\) is initial velocity, and \(t\) is time. But here we need to find the increase in velocity, and we know that \(a=\frac{\Delta v}{t}\), so \(\Delta v=a\times t\).
Given \(a = 5\space m/s^{2}\) and \(t = 2\space s\).

Step 2: Calculate the increase in velocity

Substitute the values into the formula \(\Delta v=a\times t\).
\(\Delta v=5\space m/s^{2}\times2\space s = 10\space m/s\)

Step 1: Convert initial velocity to SI units

Initial velocity \(u = 18\space km/h\). To convert to \(m/s\), use the conversion factor \(1\space km = 1000\space m\) and \(1\space h=3600\space s\). So \(u=\frac{18\times1000}{3600}=5\space m/s\), final velocity \(v = 0\space m/s\) (comes to stop), time \(t = 2\space s\).
Retardation \(r=\frac{u - v}{t}\) (retardation is negative acceleration, magnitude is \(\frac{u - v}{t}\) when \(v\lt u\)).

Step 2: Calculate retardation

Substitute \(u = 5\space m/s\), \(v = 0\space m/s\) and \(t = 2\space s\) into the formula.
\(r=\frac{5 - 0}{2}=2.5\space m/s^{2}\)

Step 1: Recall the formula for final velocity

The formula for final velocity \(v=u+at\), here \(a\) is retardation (so \(a=- 2\space m/s^{2}\) as it is decelerating), initial velocity \(u = 20\space m/s\), time \(t = 5\space s\).

Step 2: Calculate final velocity

Substitute the values into the formula \(v=u+at\).
\(v = 20+( - 2)\times5=20 - 10 = 10\space m/s\)

Answer:

The increase in velocity is \(10\space m/s\)

Second Sub - Question (Toy car retardation)