
Suppose a truck starts rest with an acceleration of \[1.5m{s^{ - 2}}\] while a car \[150m\]behind starts from rest with an acceleration of\[2m{s^{ - 2}}\].
Then find how long will it take before both the truck and car are side by side,
And also find the distance travelled by each.
Answer
513k+ views
Hint: In order to answer this question at first we must find time taken by the object during displacement and also the initial , final velocity and acceleration . Here we use the equation of motion to finding displacement i.e. \[ \Rightarrow S = ut + \dfrac{1}{2}a{t^2}\]
Complete step by step solution:
From this question we get –
Velocity of car\[ = 2\dfrac{m}{{{s^2}}}\],
Velocity of truck\[ = 1.5\dfrac{m}{{{s^2}}}\],
Therefore the relative velocity between the car and the truck will be
\[
{v_r} = (2 - 1.5)\dfrac{m}{{{s^2}}} \\
{v_r} = 0.5\dfrac{m}{{{s^2}}} \\
\]
Now let the Initial velocity \[u = 0\]
Now by applying the distance formula of motion –
\[ \Rightarrow S = ut + \dfrac{1}{2}a{t^2}\]
\[ \Rightarrow 150 = 0 + \dfrac{1}{2} \times \dfrac{5}{{10}}{t^2}\]
\[ \Rightarrow 150 = \dfrac{1}{2} \times \dfrac{5}{{10}}{t^2}\]
\[ \Rightarrow \dfrac{{3000}}{5} = {t^2}\]
\[
\Rightarrow {t^2} = 600 \\
\Rightarrow t = \sqrt {600} \\
\Rightarrow t = 24.5 \\
\]
Therefore, time take before both the truck and the car are side by side is \[24.5s\]
Again, given Uniform acceleration of truck \[{u_{truck}} = 0\]
And uniform acceleration of car \[{u_{car}} = 0\]
(Acceleration) \[{a_{truck}} = 1.5\dfrac{m}{{{s^2}}}\] and \[{a_{car}} = 2\dfrac{m}{{{s^2}}}\]
So, Distance travelled by the truck \[{S_{truck}} = ut + \dfrac{1}{2}a{t^2}\]
\[{S_{truck}} = \dfrac{1}{2} \times 1.5{(24.5)^2}\]
\[{S_t}_{ruck} = 450.19\]
So, distance travelled by car \[{S_{car}} = ut + \dfrac{1}{2}a{t^2}\]
\[{S_{car}} = \dfrac{1}{2} \times 2 \times {\left( {24.5} \right)^2}\]
\[{S_{car}} = {(24.5)^2}\]
\[{S_{car}} = 600.25\]
Note:
If there are no external forces applied on the object then its keep remains in the same state. This is defined by scientist Newton as “Law of Inertia”. If an object starts late after a vehicle then it will cross the vehicle which started first only if the velocity of the second vehicle is more that the initial vehicle.
Complete step by step solution:
From this question we get –
Velocity of car\[ = 2\dfrac{m}{{{s^2}}}\],
Velocity of truck\[ = 1.5\dfrac{m}{{{s^2}}}\],
Therefore the relative velocity between the car and the truck will be
\[
{v_r} = (2 - 1.5)\dfrac{m}{{{s^2}}} \\
{v_r} = 0.5\dfrac{m}{{{s^2}}} \\
\]
Now let the Initial velocity \[u = 0\]
Now by applying the distance formula of motion –
\[ \Rightarrow S = ut + \dfrac{1}{2}a{t^2}\]
\[ \Rightarrow 150 = 0 + \dfrac{1}{2} \times \dfrac{5}{{10}}{t^2}\]
\[ \Rightarrow 150 = \dfrac{1}{2} \times \dfrac{5}{{10}}{t^2}\]
\[ \Rightarrow \dfrac{{3000}}{5} = {t^2}\]
\[
\Rightarrow {t^2} = 600 \\
\Rightarrow t = \sqrt {600} \\
\Rightarrow t = 24.5 \\
\]
Therefore, time take before both the truck and the car are side by side is \[24.5s\]
Again, given Uniform acceleration of truck \[{u_{truck}} = 0\]
And uniform acceleration of car \[{u_{car}} = 0\]
(Acceleration) \[{a_{truck}} = 1.5\dfrac{m}{{{s^2}}}\] and \[{a_{car}} = 2\dfrac{m}{{{s^2}}}\]
So, Distance travelled by the truck \[{S_{truck}} = ut + \dfrac{1}{2}a{t^2}\]
\[{S_{truck}} = \dfrac{1}{2} \times 1.5{(24.5)^2}\]
\[{S_t}_{ruck} = 450.19\]
So, distance travelled by car \[{S_{car}} = ut + \dfrac{1}{2}a{t^2}\]
\[{S_{car}} = \dfrac{1}{2} \times 2 \times {\left( {24.5} \right)^2}\]
\[{S_{car}} = {(24.5)^2}\]
\[{S_{car}} = 600.25\]
Note:
If there are no external forces applied on the object then its keep remains in the same state. This is defined by scientist Newton as “Law of Inertia”. If an object starts late after a vehicle then it will cross the vehicle which started first only if the velocity of the second vehicle is more that the initial vehicle.
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