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 \(g = 10 \text{ m/s}^2\).)
- A body falling from rest has a velocity '\(v\)' after it falls through a distance '\(h\)'. The distance it has to fall down further, for its velocity to become double, is _____ times '\(h\)' .
- By applying the brakes without causing a skid, the driver of a car is able to stop his car within a distance of 5 m, if it is going at 36 kmph. If the car were going at 72 kmph, using the same brakes, he can stop the car over a distance of:
- A ball, initially at rest, falls freely from the top of a building and reaches a maximum velocity of \(40\text{ m/s}\). Find the height of the building. (Use acceleration due to gravity, \(g = 10\text{ m/s}^2\))
- A ball is thrown vertically upwards with initial velocity \(V_0\) and returns to its starting point in 6 seconds then the initial velocity with which the ball was thrown will be
- A ball, having speed \(V_0\) moves in a straight line under the influence of a constant acceleration a. Then its final speed after travelling a distance x for time t will be
- A ball is thrown vertically upward with a speed of 30 m/s. The magnitude of its displacement after 4 s will be _____ (Take \(g = 10\ m/s^2\).)
A particle experiences constant acceleration for 20 s after starting from rest. If it travels a distance X 1, in the first 10 s and distance X 2in the remaining 10 s, then which of the following is true?
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).
Which of the following is an equation of motion?
I. u = v + at
II. 2as = v 2– u 2
A stone is dropped from a ballon going up with a uniform velocity of 5m/sec. If the ballon was 50 m high, then the stone was dropped, the height the ballon from ground when stone hits the ground will be:
(g = 10 m/s 2)
