Question

A bus of mass 1000kg has an engine which produces a constant power of 50kW. If the resistance to the motion, assumed constant is 1000N. The maximum speed at which the bus can travel on level road and the acceleration when it is travelling at 25m/s, will respectively be:
A. $50m{{s}^{-1}},1.0m{{s}^{-2}}$
B. $1.0m{{s}^{-1}},50m{{s}^{-2}}$
C. $5.0m{{s}^{-1}},10m{{s}^{-2}}$
D. $10m{{s}^{-1}},5m{{s}^{-2}}$

Answer 1

Power generated by the engine of the bus is defined as the rate of work done with respect to time. Also, work done is the transfer of energy. Work done is calculated as the product of force and displacement. The maximum speed of the bus can be calculated by using the equation of motion and acceleration is calculated from the formula of power generated by the engine.

Formula used:
Power generated is calculated as:
$P=W/t$
Also the work done is given by:
$W=F\times d$

Complete answer:
Power is the rate of change of work done on the object. Power is the work done divided by time. It is given by:
$P=W/t$
Also, work done is defined as the force that acts on the body to displace it from one place to another. It is the product of force applied to displacement.
$W=F\times d$
By putting the value of work done in power generated, we get:
$\begin{aligned} & P=W/t \\\ & \Rightarrow P=(F\times d)/t \\\ & \Rightarrow P=F\times v \\\ \end{aligned}$
Where, v = velocity which is the rate of change of displacement with respect to time.
Now, we will multiple and divide the denominator and numerator to time. Then, the power generated is:
$\begin{aligned} & P=(F\times v\times t)/t \\\ & \Rightarrow P=F\times t\times (v/t) \\\ & \Rightarrow P=F\times a\times t \\\ \end{aligned}$
The above equation is the modified form of power generated.
Now, we will find the acceleration:
$\begin{aligned} & F=ma \\\ & \Rightarrow F/m=a \\\ & \Rightarrow a=1000N/1000kg \\\ & \Rightarrow a=1m{{s}^{-2}} \\\ \end{aligned}$
Now, the time interval from the modified formula of power generated is:
$\begin{aligned} & P=Fat \\\ & \Rightarrow t=P/Fa \\\ & \Rightarrow t=50kW/1000N\times 1m{{s}^{-2}} \\\ & \Rightarrow t=(50\times {{10}^{3}})/1000 \\\ & \Rightarrow t=50s \\\ \end{aligned}$
Now we will calculate the maximum speed by using the equation of motion, also the initial velocity is given in the question. We will find the displacement first and then speed is calculated.
$\begin{aligned} & S=ut+(1/2)a{{t}^{2}} \\\ & \Rightarrow S=(25\times 50)+(1/2)\times 1\times {{50}^{2}} \\\ & \Rightarrow S=1250+1250 \\\ & \Rightarrow S=2500m \\\ \end{aligned}$
Now finding maximum speed,
$\begin{aligned} & v=S/t \\\ & \Rightarrow v=2500/50 \\\ & \Rightarrow v=50m{{s}^{-1}} \\\ \end{aligned}$
Hence, the maximum speed and corresponding acceleration comes out to be $50m{{s}^{-1}},1.0m{{s}^{-2}}$

So option A is the correct option.

Note:
Power generated is given by the ratio of work done to time. Work done is given by product force and displacement. Using the formula of work done in power generated, we will have a modified form power generated, and we will find a time interval from this formula. We use equations of motion to find the displacement, and further we use it to find maximum speed at which bus travels.