Calculating Distance Travelled Under Constant Acceleration
To find the distance travelled by the car, we can use the equations of motion for constant acceleration.
Given Information:
- Initial velocity, \(u = 0 \text{ m/s}\) (since the car starts from rest)
- Acceleration, \(a = 3 \text{ m/s}^2\)
- Time, \(t = 10 \text{ s}\)
Relevant Formula:
The second equation of motion relates distance (\(s\)), initial velocity (\(u\)), acceleration (\(a\)), and time (\(t\)):
\(s = ut + \frac{1}{2}at^2\)
Calculation Steps:
- Substitute the known values into the formula:
\(s = (0 \text{ m/s})(10 \text{ s}) + \frac{1}{2}(3 \text{ m/s}^2)(10 \text{ s})^2\)
- Simplify the terms:
The first term (\(ut\)) becomes $0$. The square of time is \((10 \text{ s})^2 = 100 \text{ s}^2\).
\(s = 0 + \frac{1}{2}(3 \text{ m/s}^2)(100 \text{ s}^2)\)
- Calculate the final value:
\(s = \frac{1}{2}(300 \text{ m})\)
\(s = 150 \text{ m}\)
Result:
The distance travelled by the car in \(10 \text{ s}\) is \(150 \text{ m}\).
Similar Questions
- A ball thrown up vertically returns to the ground after 10 second. Find the velocity with which it was thrown up? (if g = 10 m/s2).
- A ball is thrown vertically upward with a speed of 30 m/s. The magnitude of its displacement after 4 s will be ______ (Take \(g = 10 \text{ m/s}^2\).)
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Important Questions from Kinematic equations for uniformly accelerated motion
Which of the following is an equation of motion?
I. u = v + at
II. 2as = v 2– u 2
- A ball is thrown vertically upward with a speed of 40 m/s. The time taken by the ball to reach the maximum height would be approximately
- A ball thrown up vertically returns to the ground after 10 second. Find the velocity with which it was thrown up? (if g = 10 m/s2).
If the distance travelled by a body in the $n^{th}$ second is given by $(7 + 5n)$ m, then find the initial velocity and acceleration of the body respectively.
Which of the following is an equation of motion?
I. u = v + at
II. 2as = v 2– u 2
