Question

Consider a car moving with constant acceleration along a straight road and the distance covered by the car is given by equation s = 5t² + 3t + 9 meters. Then find the ratio of acceleration and initial velocity of the car at the start.

1. 10 ∶ 3
2. 1 ∶ 2
3. 3 ∶ 10
4. 2 ∶ 1

Answer 1

Correct Answer - Option 1 : 10 ∶ 3

CONCEPT:

• The velocity of the particle is defined as the rate of change of its displacement
• And can be expressed as

\(\vec v = \frac{{\overrightarrow {dx} }}{{dt}}\)

​Equation of Kinematics:

• These are the various relations between u, v, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
• Equations of motion can be written as

⇒ v = u + at

\(⇒ s=ut+\frac{1}{2}at^{2}\)

⇒ v2 = u2+ 2as

where, u = Initial velocity, v = Final velocity, g = Acceleration due to gravity, t = time, and h = height/Distance covered

CALCULATION:

Given that- Distance traveled by a car (s) = 5t2 + 3t + 9

So if we differentiate the given equation with respect to t, we will get final velocity at any instant of time and it can be given as

\(\Rightarrow \frac{{ds}}{{dt}} = 5 \times 2t + 3 = 3 + 10t\)

\(\Rightarrow v = 3 + 10t\;\;\;\;\;\;\left( \because {\frac{{ds}}{{dt}} = v} \right)\)     ---(1)

• Kinematic equation

⇒ v = u + at    ---(2)

• Now comparing this equation with the kinematic equation, we can see that initial velocity, u = 3 m/s, and acceleration, a = 10 m/s2
• The ratio of acceleration and initial velocity of the car at the start will be

\(\Rightarrow \frac{a}{v}=\frac{10}{3}\)