Question

A motor car moving at a speed of 72kmph cannot come to a stop in less than 3.0s while for a truck this time interval is 5.0s. On a highway the car is behind the truck both moving at $72\dfrac{{km}}{h}$. The truck gives a signal that it is going to stop at an emergency. At what distance the car should be from the truck so that it does not bump into (collide with) the truck. Human response time is 0.5s:
A) 1.25m.
B) 1.5m.
C) 1m.
D) 0.5m.

Answer 1

Hint:The reaction time is the time in which the human being takes to react in certain conditions. Newton's law of motion relations can be used to solve this problem as the acceleration on the car and the truck is constant.

Formula used:
The formula of the first relation of the Newton’s law of motion is given by,
$ \Rightarrow v = u + at$
Where final velocity is v, the initial velocity is u the acceleration is a and the time taken is t.
The formula of the second relation of the Newton’s law of motion is given by,
$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
Where displacement is s the initial velocity is u the time taken is t and the acceleration is a.

Complete step by step solution:
It is given in the problem that a motor car moving at a speed of $72\dfrac{{km}}{h}$ cannot come to a stop in less than 3.0s while for a truck this time interval is 5.0s on a highway the car is behind the truck both moving at$72\dfrac{{km}}{h}$ the truck gives a signal that it is going to stop at emergency we need to tell the distance at which the car should be from the truck so that it does not bump into (collide with) the truck if human response time is 0.5s.
Converting the speed of truck and car from $\dfrac{{km}}{h}$into$\dfrac{m}{{\sec }}$.
$ \Rightarrow {u_t} = {u_c} = 72 \times \dfrac{5}{{18}} = 20\dfrac{m}{s}$
For motion of the truck.
Let’s calculate the value of the acceleration of the truck.
The formula of the first relation of the Newton’s law of motion is given by,
$ \Rightarrow v = u + at$
Where final velocity is v, the initial velocity is u the acceleration is a and the time taken is t.
The final velocity of the truck is zero and the time taken by the truck to stop is 5sec.
$ \Rightarrow v = u + at$
$ \Rightarrow 0 = 20 + {a_t} \times \left( 5 \right)$
$ \Rightarrow {a_t} = \dfrac{{ - 20}}{5}$
$ \Rightarrow {a_t} = - 4\left( {\dfrac{m}{{{s^2}}}} \right)$
For motion of car,
Let’s calculate the value of acceleration of the car.
The formula of the first relation of the Newton’s law of motion is given by,
$ \Rightarrow v = u + at$
Where final velocity is v, the initial velocity is u the acceleration is a and the time taken is t.
The final velocity of the car will be zero as it stops in order to resist the collision and the time taken is 3 sec.
$ \Rightarrow v = u + at$
$ \Rightarrow 0 = 20 + {a_c} \times \left( 3 \right)$
$ \Rightarrow {a_c} = \dfrac{{ - 20}}{3}\left( {\dfrac{m}{{{s^2}}}} \right)$
Let the time taken by the truck to stop is t sec.
The velocity of the truck after t sec. is equal to,
$ \Rightarrow v = u + at$
$ \Rightarrow {v_t} = 20 - 4t$
The velocity of the car after $t - 0 \cdot 5$ sec. is equal to,
$ \Rightarrow v = u + at$
$ \Rightarrow {v_c} = 20 - \left( {\dfrac{{20}}{3}} \right)\left( {t - 0 \cdot 5} \right)$
The time taken is $t - 0 \cdot 5$ sec. because the human response time is 0.5 sec.
The velocity of the truck and car should be equal in order to avoid collision.
$ \Rightarrow {v_t} = {v_c}$
$ \Rightarrow 20 - 4t = 20 - \left( {\dfrac{{20}}{3}} \right)\left( {t - 0 \cdot 5} \right)$
$ \Rightarrow 4t = \left( {\dfrac{{20}}{3}} \right)\left( {t - 0 \cdot 5} \right)$
$ \Rightarrow 12t = 20t - 10$
$ \Rightarrow t = \dfrac{{10}}{8}$
$ \Rightarrow t = \dfrac{5}{4}s$
Let us calculate the distance travelled by the truck in t sec.
The formula of the second relation of the Newton’s law of motion is given by,
$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
Where displacement is s the initial velocity is u the time taken is t and the acceleration is a.
Initial velocity is $u = 20\dfrac{m}{s}$ the time taken is $t = \dfrac{5}{4}s$ the acceleration of the truck is${a_t} = - 4\left( {\dfrac{m}{{{s^2}}}} \right)$ we get.
$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
$ \Rightarrow {s_t} = 20 \times \dfrac{5}{4} - \dfrac{1}{2} \times 4 \times {\left( {\dfrac{5}{4}} \right)^2}$
$ \Rightarrow {s_t} = 25 - 2 \times \dfrac{{25}}{{16}}$
$ \Rightarrow {s_t} = 25 - \dfrac{{25}}{8}$
$ \Rightarrow {s_t} = 25 - 3 \cdot 125$
$ \Rightarrow {s_t} = 21 \cdot 875m$………eq. (1)
Let us calculate the distance travelled by the car in t sec.
The formula of the second relation of the Newton’s law of motion is given by,
$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
Where displacement is s the initial velocity is u the time taken is t and the acceleration is a.
The initial velocity of the car is $u = 20\dfrac{m}{s}$ the time taken is $t = \left( {\dfrac{5}{4} - 0 \cdot 5} \right)s = 0 \cdot 75s$and the acceleration of the car is $ \Rightarrow {a_c} = \dfrac{{ - 20}}{3}\left( {\dfrac{m}{{{s^2}}}} \right)$ we get.
$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
$ \Rightarrow {s_c} = 20 \times \dfrac{5}{4} - \dfrac{1}{2} \times \dfrac{{20}}{3} \times {\left( {0 \cdot 75} \right)^2}$
$ \Rightarrow {s_c} = 25 - \dfrac{{10}}{3} \times {\left( {0 \cdot 75} \right)^2}$
$ \Rightarrow {s_c} = 25 - 1 \cdot 875$
$ \Rightarrow {s_c} = 23 \cdot 125m$………eq. (2)
The difference between the distance travelled by the car and the truck is equal to,
$ \Rightarrow {s_c} - {s_t}$
$ \Rightarrow {s_c} - {s_t} = 23.125 - 21 \cdot 875$
$ \Rightarrow {s_c} - {s_t} = 1 \cdot 25m$
So the difference between the car and the truck should be 1.25 ms in order to avoid the collision.

The correct answer for this problem is option A.

Note: It is advisable for students to remember the formula of the relations of Newton's law of motion as it is very helpful in solving problems like these. The time taken by the truck is $t$ and the time taken by the car should be $t - 0 \cdot 5$ to calculate motion because the time taken for response time is 0.5sec.