Question

If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of :

• A. 2880 m
• B. 1440 m
• C. 400 m
• D. 20 m

Answer 1

Answer: (C)

Key Idea: Distance covered will be according to 2nd equation of motion i. e., $s=ut+\frac{1}{2}a{t}^2$ . Here, $u=0,t=20s$ $v=144km/h$ $=144\times \frac{5}{18}m/s$ $=40m/s$ From 1st equation of motion, $v=u+at$ $\Rightarrow$ $a=\frac{v-u}{t}=\frac{40-0}{20}=2m/s^2$ Now, 2nd equation of motion, $s=ut+\frac{1}{2}a{t}^2$ $=0+\frac{1}{2}\times 2\times {(20)}^2=400m$ NOTE: If the car decelerates, the acceleration will be negative and is termed as retardation.