Question

A motorcycle and a car start from rest from the same place at the same time and travel in the same direction. The motorcycle accelerates at $1.0\;m{s^{ - 1}}$ up to a speed of $36\;km{h^{ - 1}}$ and the car at $0.5\;m{s^1}$ up to a speed of $54\;km{h^{ - 1}}$. Find the time at which the car would overtake the motorcycle?
A. $20\;s$
B. $25\;s$
C. $30\;s$
D. $35\;s$

Answer 1

Hint: The motorcycle and car start from the same place, at same time and move in the same direction. But they accelerate at different accelerations and also the maximum speed is varied. At one point the car will overtake the motorcycle, at that point the distance travelled by the car and motorcycle are the same. Hence, by finding the distance travelled by the car and motorcycle relation, the time when the car overtakes the motorcycle will be calculated.

Useful formula:
The equations of motion,
$v = u + at$
$S = ut + \dfrac{1}{2}a{t^2}$
Where, $v$ is the final velocity of the body, $u$ is the initial velocity of the body, $a$ is the acceleration of the body, $t$ is the time taken to travel and $S$ is the distance travelled by the body.

Speed = Distance *Time.

Given data:
Acceleration of the motorcycle, ${a_m} = 1\;m{s^{ - 2}}$
Acceleration of the car, ${a_c} = 0.5\;m{s^{ - 2}}$
Initial velocity of the motorcycle, ${u_m} = 0\;km{h^{ - 1}}$
Initial velocity of the car, ${u_c} = 0\;km{h^{ - 1}}$
Final velocity of the motorcycle, ${v_m} = 36\;km{h^{ - 1}} = 10\;m{s^{ - 1}}$
Final velocity of the car, ${v_c} = 54\;km{h^{ - 1}} = 15\;m{s^{ - 1}}$

Step by step solution:
Applying equation of motion on motorcycle,
${v_m} = {u_m} + {a_m}{t_1}\;......................................\left( 1 \right)$
Where, ${u_m}$ is the initial speed of motorcycle and ${t_1}$ is the time taken by motorcycle to reach maximum speed.
Substituting given values in equation (1),
$10\;m{s^{ - 1}} = 0 + \left( {1\;m{s^{ - 2}} \times {t_1}} \right) \\\ {t_1} = 10\;s \\\$
The distance covered by motorcycle to attain maximum speed,
${S_m} = {u_m}{t_1} + \dfrac{1}{2}{a_m}{t_1}^2$
Substitute the given values, we get
${S_m} = \left( {0 \times {t_1}} \right) + \dfrac{1}{2}\left( {1\;m{s^{ - 2}}} \right){\left( {10} \right)^2} \\\ {S_m} = \dfrac{{100}}{2}\;m \\\ {S_m} = 50\;m \\\$
The time duration of the motorcycle with maximum speed before overtaking is $\left( {t - 10} \right)\;s$.
The distance covered by the motorcycle with maximum speed before overtaking,
${S_m}^\prime = {v_m} \times \left( {t - 10} \right)$
Substitute the given values in above equation, we get
${S_m}^\prime = 10\;m{s^{ - 1}} \times \left( {t - 10} \right) \\\ {S_m}^\prime = \left( {10t - 100} \right)\;m \\\$
Hence, the total distance travelled by motorcycle, $S = {S_m} + {S_m}^\prime$
Substitute the values of distance in above relation,
$S = 50\;m + \left( {10t - 100} \right)\;m \\\ S = \left( {10t - 50} \right)\;m\;.................................\left( 2 \right) \\\$

Applying equation of motion on car,
${v_c} = {u_c} + {a_c}{t_2}\;......................................\left( 3 \right)$
Where, ${u_c}$ is the initial speed of the car and ${t_2}$ is the time taken by car to reach maximum speed.
Substituting given values in equation (3),
$15\;m{s^{ - 1}} = 0 + \left( {0.5\;m{s^{ - 2}} \times {t_2}} \right) \\\ {t_2} = 30\;s \\\$
The distance covered by car to attain maximum speed,
${S_c} = {u_c}{t_2} + \dfrac{1}{2}{a_c}{t_2}^2$
Substitute the given values, we get
${S_c} = \left( {0 \times {t_2}} \right) + \dfrac{1}{2}\left( {0.5\;m{s^{ - 2}}} \right){\left( {30} \right)^2} \\\ {S_c} = \dfrac{{450}}{2}\;m \\\ {S_c} = 225\;m \\\$
The time duration of the car with maximum speed before overtaking is $\left( {t - 30} \right)\;s$.
The distance covered by the car with maximum speed before overtaking,
${S_c}^\prime = {v_c} \times \left( {t - 30} \right)$
Substitute the given values in above equation, we get
${S_c}^\prime = 15\;m{s^{ - 1}} \times \left( {t - 30} \right) \\\ {S_c}^\prime = \left( {15t - 450} \right)\;m \\\$
Hence, the total distance travelled by car, $S = {S_c} + {S_c}^\prime$
Substitute the values of distance in above relation,
$S = 225\;m + \left( {15t - 450} \right)\;m \\\ S = \left( {15t - 225} \right)\;m\;.................................\left( 4 \right) \\\$

Since, the distance travelled by motorcycle and car is the same. Then equating equation (2) and (4),
$\left( {10t - 50} \right)\;m = \left( {15t - 225} \right)\;m \\\ 15t - 10t = 225 - 50 \\\ 5t = 175 \\\ t = \dfrac{{175}}{5}\;s \\\ t = 35\;s \\\$
Hence, the option (D) is correct.

Note: The motorcycle and the car start at the same point and at same time and move in the same direction. The motorcycle accelerates with more acceleration compared to a car. But the maximum speed of the motorcycle is lesser than the car. Hence, the car will overtake the motorcycle at one point.