A) 2880 m
B) 1440 m
C) 400 m
D) 20 m
Correct Answer: C
Solution :
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.
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