Question

A body travels in a straight line according to the equation $x=4+3t-t^2$. The ratio of distance travelled to displacement in t=0 to t= 2 sec. is

• A. 5:4
• B. 5:2
• C. 1
• D. 13:6

Answer 1

The correct option is A 5:4

Given $x=4+3t-t^2$,

Velocity $v=\frac{dx}{dt}=3-2t$,

Plotting the graph v-t as shown in the figure.

Area of velocity-time diagram gives the displacement.

Displacement $=\frac{1}{2}\times 3\times 1.5-\frac{1}{2}\times 1\times \frac{1}{2}=2$

Distance $=\frac{1}{2}\times 3\times 1.5+\frac{1}{2}\times 1\times \frac{1}{2}=\frac{10}{4}$

Ratio of distance to displacement$=\frac{\frac{10}{4}}{2}=\frac{5}{4}$ or 5:4