Velocity-time graph of a particle is given as :
then average speed of particle from t = 0 to t = 5 sec is :
Area bounded with Velocity time graph represents displacement. Magnitude of Area bounded by Velocity time graph represents distance.
Total Area = 60 so, distance = 60 m and time = 5 sec.
Average speed = 12 m/s
A body moving with a uniform acceleration crosses a distance of 65 m in the 5th second and 105 m in 9th second. How far will it go in first 20 s?
We will use the formula for the distance covered in the \(n\)-th second:
\[
s_n = u + \frac{a}{2}(2n - 1)
\]
Given:
- Distance in the 5th second: \( s_5 = 65 \, \text{m} \)
- Distance in the 9th second: \( s_9 = 105 \, \text{m} \)
Using the formula for both:
1. \( s_5 = u + \frac{a}{2}(9) = 65 \)
\[
u + \frac{9a}{2} = 65 \tag{1}
\]
2. \( s_9 = u + \frac{a}{2}(17) = 105 \)
\[
u + \frac{17a}{2} = 105 \tag{2}
\]
Now, subtract equation (1) from equation (2):
\[
\left( u + \frac{17a}{2} \right) - \left( u + \frac{9a}{2} \right) = 105 - 65
\]
\[
\frac{8a}{2} = 40 \implies 4a = 40 \implies a = 10 \, \text{m/s}^2
\]
Substitute \( a = 10 \) into equation (1):
\[
u + \frac{9 \times 10}{2} = 65 \implies u + 45 = 65 \implies u = 20 \, \text{m/s}
\]
Now, to find the total distance covered in 20 seconds:
\[
S = ut + \frac{1}{2} a t^2 = 20 \times 20 + \frac{1}{2} \times 10 \times 20^2
\]
\[
S = 400 + 0.5 \times 10 \times 400 = 400 + 2000 = 2400 \, \text{m}
\]
Thus, the body will cover 2400 meters in 20 seconds.
A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5th second to that covered in 5 s in
For a body starting from rest with constant acceleration:
- Distance covered in the 5th second: \( s_5 = u + \frac{a}{2}(2n - 1) \), where \( u = 0 \) and \( n = 5 \):
\[
s_5 = \frac{a}{2} (9) = \frac{9a}{2}
\]
- Distance covered in 5 seconds: \( S = \frac{1}{2} a t^2 \), where \( t = 5 \):
\[
S = \frac{1}{2} a (5)^2 = \frac{25a}{2}
\]
Ratio of distances:
\[
\frac{s_5}{S} = \frac{\frac{9a}{2}}{\frac{25a}{2}} = \frac{9}{25}
\]
Thus, the ratio is 9:25.
A car is travelling at 72 kmh–¹ and is 20 m from a barrier when the driver puts on the brakes. The car hits the barrier 2s later. What is the magnitude of the constant deceleration?
Given:
- Initial velocity: \( u = 20 \, \text{m/s} \)
- Time: \( t = 2 \, \text{seconds} \)
- Distance to the barrier: \( s = 20 \, \text{m} \)
Using the equation of motion:
\[
s = ut + \frac{1}{2} a t^2
\]
\[
20 = 20 \times 2 + \frac{1}{2} a \times 2^2
\]
\[
20 = 40 + 2a
\]
\[
2a = -20
\]
\[
a = -10 \, \text{m/s}^2
\]
Thus, the magnitude of the deceleration is 10 m/s².
The acceleration a of a particle starting from rest varies with time according to relation a = αt + β. The velocity of the particle after a time t will be
Given acceleration:
\[
a = \alpha t + \beta
\]
The velocity after time \(t\) is:
\[
v(t) = \frac{\alpha t^2}{2} + \beta t
\]
The position of a particle x (in meters) at a time t second is given by the relation \[r=\left( 3t\hat{i}-t^{2}\hat{j}+4\hat{k} \right)\] . Calculate the magnitude of velocity of the particular after 5 s.
Given the position vector:
\[
\mathbf{r} = (3t)\hat{i} - (t^2)\hat{j} + 4\hat{k}
\]
To find the velocity, differentiate the position vector with respect to time \(t\):
\[
\mathbf{v} = \frac{d\mathbf{r}}{dt} = \left( \frac{d}{dt}(3t) \hat{i} + \frac{d}{dt}(-t^2) \hat{j} + \frac{d}{dt}(4) \hat{k} \right)
\]
Calculating the derivatives:
\[
\mathbf{v} = (3)\hat{i} - (2t)\hat{j} + (0)\hat{k} = 3\hat{i} - 2t\hat{j}
\]
Now, substitute \(t = 5\) s:
\[
\mathbf{v}(5) = 3\hat{i} - 2(5)\hat{j} = 3\hat{i} - 10\hat{j}
\]
Calculate the magnitude of the velocity:
\[
|\mathbf{v}| = \sqrt{(3)^2 + (-10)^2} = \sqrt{9 + 100} = \sqrt{109}
\]
Thus, the magnitude of velocity after 5 seconds is: 10.44 m/s
The velocity of particle is \[v=v_{0}+gt+ft^{2}\]. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is
Given the velocity function:
\[
v = v_0 + gt + ft^2
\]
We can find displacement by integrating the velocity function with respect to time. The displacement \(x(t)\) is the integral of velocity:
\[
x(t) = \int (v_0 + gt + ft^2) \, dt
\]
Integrating:
\[
x(t) = v_0 t + \frac{gt^2}{2} + \frac{ft^3}{3} + C
\]
Since \(x = 0\) at \(t = 0\), we know \(C = 0\). Therefore, the position function is:
\[
x(t) = v_0 t + \frac{gt^2}{2} + \frac{ft^3}{3}
\]
Now, for \(t = 1\):
\[
x(1) = v_0(1) + \frac{g(1)^2}{2} + \frac{f(1)^3}{3}
\]
Simplifying:
\[
x(1) = v_0 + \frac{g}{2} + \frac{f}{3}
\]
Thus, the displacement after unit time \(t = 1\) is:
\[
x(1) = v_0 + \frac{g}{2} + \frac{f}{3}
\]
A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity v that varies as v = α√x . The displacement of the particle varies with time as
Given that the velocity \(v = \alpha \sqrt{x}\), we need to find the displacement \(x\) as a function of time \(t\).
1. \( v = \frac{dx}{dt} = \alpha \sqrt{x} \)
2. Rearranging and separating variables:
\[
\frac{dx}{\sqrt{x}} = \alpha \, dt
\]
3. Integrate both sides:
\[
2\sqrt{x} = \alpha t
\]
4. Solving for (x):
\[ x = \frac{\alpha^2 t^2}{4} \]
Thus, the displacement of the particle varies with time as \(x = \frac{\alpha^2 t^2}{4}\).
The v-t graph for a particle is as shown. The distance travelled in the first four second is
Area bounded by v-t graph represents displacement. Here.
Area = (1/2)×4×8= 16
So, Distance travelled is 16 m.
A particle starts form rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be
Area of acceleration time graph represents change in velocity. As acceleration is always positive speed is always increasing.
Area = (1/2)× 10× 11 = 55
So, max. speed of 55 m/s