Question

1 point
Q1. A body moving with an
acceleration of 5 m/s2 started
with yriocity of 10 m/s. What
will be the distance travelled in
10 seconds ?
150 m
ОО
250 m
350 m
400 m​

Answer 1

Answer

• Distance travelled will be 350 m [Option C]

Explanation

Given

• Acceleration of a body is 5 m/s²
• Initial Velocity = 10 m/s
• Time = 10 sec

To Find

• The distance travelled during this time

Solution

• We shall here use the second equation of motion as we are given the initial velocity, the time taken and the acceleration of the body. Here we may think of using the third equation of motion but it won't work as we are not given the final velocity of the body

✭ Distance Covered

→ s = ut + ½at²

→ s = 10×10 + ½ × 5 × 10²

→ s = 100 + ½ × 5 × 100

→ s = 100 + 5×50

→ s = 100 + 250

→ s = 350 m

∴ Our answer here is Option C

Know More

Other equations of motion

• v = u+at [First equation of motion]
• v²-u² = 2as [Third Equation of motion]

Answer 2

Answer:

$\huge \sf \: given$

• Acceleration of body = 5 m/s²
• Velocity of the body = 10 m/s
• Time taken = 10 sec

$\huge \sf \: to \: find$

Distance travelled by body

$\huge \sf \: solution$

So, we have to find distance travelled. So we will use the second equation of motion.

$\huge \bf \: s \: = ut \: + \frac{1}{2} \times {at}^{2}$

Here,

S - Distance travelled

U - Velocity

T - Time

A - Acceleration

Now, according to Question

$\sf \implies \: 10 \times 10 + \frac{1}{2} \times {5 \times 10}^{2}$

$\sf \implies \: 100 + \frac{1}{2} \times 5 \times 100$

$\sf \implies \: 100 + 5 \times 50$

$\sf \implies100 + 250$

$\huge \fbox {option \: c = 350 \: m}$