The displacement x of a particle along x axis is given by x=3+8t+7t square obtain velocity and accleration at t=19
Answers
Answer :
- Velocity of the of the particle with displacement, x = 3 + 8t + 7t² at t = 19 s, dv = 274 m/s
- Acceleration of the particle with displacement, x = 3 + 8t + 7t² is 14 m/s²
Explanation :
Given :
- Displacement of the particle, x = 3 + 7t + 8t²
- Time instant, t = 19 s
To find :
- Velocity of the particle, v = ?
- Acceleration of the particle, a = ?
Knowledge required :
- If we differentiate the displacement of a particle we will get the velocity of the particle.
- If we differentiate the velocity of a particle we will get the acceleration of that particle.
- Power rule of differentiation : x^n = n·x^(n - 1).
- Differentiation of a constant term is zero (0).
Solution :
⠀⠀⠀⠀⠀⠀⠀⠀⠀Velocity of the particle :
We know that by differentiating the displacement of a particle we will get the velocity of the particle.
So let's Differentiate the displacement of that particle with respect to t :
⠀⠀⠀=> v = d(x)/dt
⠀⠀⠀=> v = d(x)/dt = d(3 + 8t + 7t²)/dt
⠀⠀⠀=> v = d(x)/dt = d(3)/dt + d(8t)/dt + d(7t²)/dt
⠀⠀⠀=> v = d(x)/dt = 0 + d(8t¹)/dt + d(7t²)/dt
⠀⠀⠀=> v = d(x)/dt = d(8t¹)/dt + d(7t²)/dt
By applying the power rule, we get :
⠀⠀⠀=> v = d(x)/dt = 1 × 8t¹ ⁻ ¹ + 2 × 7t² ⁻ ¹
⠀⠀⠀=> v = d(x)/dt = 8 + 14t
⠀⠀⠀∴ d(v) = 8 + 14t
Hence the velocity of the particle is 8 + 7t.
Now let's find the velocity of the particle at the given instant, (i.e, t = 19 s)
⠀⠀⠀=> d(v) = 8 + 7t
⠀⠀⠀=> v_(t = 19 s) = 8 + 7t
⠀⠀⠀=> v_(t = 19 s) = 8 + 7(19)
⠀⠀⠀=> v_(t = 19 s) = 8 + 266
⠀⠀⠀=> v_(t = 19 s) = 274
⠀⠀⠀∴ v_(t = 19 s) = 274 m/s
⠀⠀⠀⠀⠀⠀Accelaration of the particle :
We know that by differentiating the velocity of a particle we will get the qcceleration of the particle.
So let's Differentiate the velocity (dv) of that particle with respect to t :
⠀⠀⠀=> a = d(v)/dt
⠀⠀⠀=> a = d(v)/dt = d(8 + 14t)/dt
⠀⠀⠀=> a = d(v)/dt = d(8)/dt + d(14t)/dt
⠀⠀⠀=> a = d(v)/dt = 0 + d(14t)/dt
⠀⠀⠀=> a = d(v)/dt = d(14t)/dt
By applying the power rule, we get :
⠀⠀⠀=> a = d(v)/dt = 1 × 14t¹ ⁻ ¹
⠀⠀⠀=> a = d(v)/dt = 14t⁰
⠀⠀⠀=> a = d(v)/dt = 14
⠀⠀⠀∴ d(a) = 14 m/s²
Therefore,
- Velocity of the particle (at t = 19s), v = 274 m/s.
- Acceleration of the particle, a = 14 m/s²
Answer:
Answer :
The maximum hieght reached by the ball is 45 m.
Explanation :
Given :
Angle of projection, θ = 30°
Intial velocity of the ball, u = 60 m/s
Final velocity of the ball, v = 0 m/s
[Final velocity of a body at maximum height is 0, v = 0]
Acceleration due to gravity, g = 10 m/s² (Approx.)
[Acceleration due to gravity = 9.8 m/s²]
To find :
Maximum height reached by the ball, h = ?
Knowledge required :
Formula for maximum height in case of a projectile :
\boxed{\sf{h_{(max.)} = \dfrac{u^{2}sin^{2}\theta}{2g}}}
h
(max.)
=
2g
u
2
sin
2
θ
Where,
h = Maximum height
u = Initial velocity
θ = Angle of Projection
g = Acceleration due to gravity
Solution :
By using the formula for maximum height in case of a projectile and substituting the values in it, we get :
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{u^{2}sin^{2}\theta}{2g}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
2g
u
2
sin
2
θ
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times sin^{2}30^{\circ}}{2 \times 10}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
2×10
60
2
×sin
2
30
∘
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times \bigg(\dfrac{1}{2}\bigg)^{2}}{20}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20
60
2
×(
2
1
)
2
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times \dfrac{1}{4}}{20}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20
60
2
×
4
1
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{3600 \times \dfrac{1}{4}}{20}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20
3600×
4
1
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{3600}{20 \times 4}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20×4
3600
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{3600}{80}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
80
3600
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{360}{8}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
8
360
\begin{gathered}:\implies \sf{h_{(max.)} = 45} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=45
\begin{gathered}\boxed{\therefore \sf{h_{(max.)} = 45\:m}} \\ \\ \\ \end{gathered}
∴h
(max.)
=45m
Therefore,
Maximum height reached by the ball, h = 45 m.